Teach Yourself Trigonometry, Revised edition

€

trigonometry .l,"#"J.l*""o and apply сore princip|es

. exp|ore the subjeсt in depth . fill the gaps in your knowledge

teaсt.f

yourself

trigonometry p. abbott revised by hugh neill

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й.h

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prcfaGG 01

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historical backgruund

1

introduсtion

2

what is trigonometцp

3

the origins of figonometry

3

the tапgent

5

intoduсtion

6

FIrst pub|ishеd in Us 1992 by сoпtemporafy Bo.oks,а Чivision iЁ.аеntiаt ptaza,iзo вast itanoo-tptr strеet, сhicago, |L 60601 UsA.

the idea of the tаngent ratio

7

bйй;.;;,

а definition 0f tаngent

I

тhis edition published 2003.

values of the tаngent

ol H0dder tleadline Ltd. Тhe tвaсh yourselt name is a registered trade mark

notation for angles and sides

Bitish Libnrу Cahloguing in Publication Data: i сйogus rЪоord tй tris title is avai|ab|e from the British шbrary.

t2

Ltbnry of Congres Cablщ Card Number: oп file. Еuston R0ad, Londoп, NW1 3BH. FIrst pub|lshed in Uк 1992 by нodder Amold, 338

of the MсGraw-Hi|l

t|ugh Neil| сopyright @ 1970, 1 991 , 1998, 2003, Paul Abbot and permitted.use. under UK сopyright |aW' п0 part /л UK Al| riqhts reserved. Apart from any

means' ue reproсuсed or trаnsmitted in аny form or by any inо|Ldiпg photocopy,.reсording or any informatioп' st.rage 0r under liсeпсe ffi;;t'i;';l syйй, йitЁoш рJiйisiion iп йiiting from fie.publisher of suсh Iicenсes (tor t'ii" iЁibЪpyпgъt,iicensiпd liilnЬi ii'iteо. Ёurther detai|s ght Liсensin g Agenсy сoрyri ъйioouction1 йaй ьi Ььtaineo

Ьiiйi.' Ь,Б"пйi.n

йiy

;ieiffiй;iйhani6a|,

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of 90 Тottenham сюurt Road' London

0:t

-floP.thе

stаtes с0pyright.Aсt of permitted.under th9. ,n Us: A|l rights reserved. Except as чlleid in any form 0r by aпy iЬi6' i'Б pi't .t йi. puьпсation may.ьe reproduсed or distributedprior written permissigп without the means. or stored in a datanase.oi-r|йevalsystem, 0f сontemporary B00ks.

а

сйii

.grЬwп HodderHead|iпe'spo|iсyistousepaрersthatarenatura|,.renewab|eandreсyсIab|e in.susчiч!]e.j.9.l9sts. тhе |ogging and йаrie йm ;йi,1.t' proсesses йЁйo ю conform t0 the rпviroпmenhl requ|ations of

й

manufaсturing

the сountry 0f origin.

lmpression

Year

wй

',.

number 10 I

876543

2010 2009 2008 2007 2006 2005 2004

04

I 10

using tangeпts

10

opposite аnd adjaсent sides

14

sanc and Gosine

18

introduсtion

19

defiпition of sine aпd сosine

20

using

Wl l 4Ll,.

Typeset by Pantek Art Ltd, мaidstone' кent ivision of Hodder tleadliпe, 338 Еuston вritaiп fш нoJоei дiпolо, ЁiЬiЬ-о. i-.i irйi, loпo* шw1 зBtl, by сox & Wlmаn Цd, Reading, Berkshire.

o э r+ o

+.lt

fie

sine аnd сosine

21

rigonometriс ratios 0f 45", 30. and 60.

25

using the ca|cu|аtor aссurately

27

slope and gradient

27

projeсti0ns

28

multistage problems

30

in ttrree dimensions

t5

introduсti0n

36

pyramid problems

36

box problems

39

wedge problems

41

аngles of any magпitude introduсtion

sine and сosine for any ang|e graphs of sine and cosine funсtions the tangent of anY angle graph of the tangent funсtion

46 48

5l 53 54 57

59

the sine aпd сosine formulae

в2

the sine formula for a triangle the ambiguous сase the сosine formu|a for a triаngIe introduction to surveying

66 68 69 73

finding ttre height of a distant objeсt

73

distanсe of an inaсcessib|r objeсt distanсe between two inaссessible but visib|e objесts triangulation

radiаns

сonverting from radians to degreеs

118 't19

the seсond set of faсtor formulаe

122

сirc|es rвlаted to a riang|e

1zfi

the сirсumсirс|e

127

the inсirсle

131

heron's formula:

132

fie

area of a triangle

134

general solution of eqшations

138

the equation siп 0 = sin the equаtioп сo{i 0 = cos

139

с

the equation taп 0 = tan

d

с

summаry of results

141 141

142

151 16:!

86

91

96

mu|tip|e aпglе formulae

the first set of faсtor formuIаe

indeх

ratios of сompound апgles

worked еxamples

114

the faсtor fomшlаe

aпswG1s

91

-0 formulae for сos(/ + 0 and cos(я - 0 formulae for tan(,4 + 0 and tan(l - 81

111

using the altemative form

&l

seоant, сoseсant and сotangent

and sin(A

111

thеformy=аsinx+Dсosx

the ecirс|es

14

110

introduсtion

76

90

$

x

108

145

relations between the rаtios

formulaе for sin (A +

0 Gos

1tl8

87

сompound angles

x+

summarу d tsigonomeшiс formulae

arеа of a сirсular seсt0r

introduсtion

а siп

76

84

length of a сirоuIar аrс

fom

75

84

radians

the

106

glo$sary

84

introduсti0n

13

63 63

areа of a triangle

12

54

solving equations iпvo|ving tangents notаtion

11

50

solving simple equаtions

so|ving equations involving сosinеs

10

more trigonometriс equatioпs

52

solving equations involving sines

09

identitiеs

46

sine, сosinе аnd tаngent

introduсtion

08

45

97 97 99 100 101

103

Г"Ц

ll

т' -

o {r q) o o

Ъaсh Yourself Trigonomеtry has bееn suЬstantially revisеd аnd ,.-ii...', to takе aЪ.oo,,t of modеrn nееds and reсеnt devеlоpmеnts in thе subjесt.

will havе aссеss to a sсiеntifiс sines, соsinеs and tangеnts' and thеir й.,,",. It is also important that thе сalсulator has a memory ;;.;й;i";"'mеdiatе?еsults сan bе stored aс|urately. No supЬЬ,t h"' been givеn about how to use thе сalсulator,-еxсеptin in thе iЁ;;;-*;ial terms. Calсulators vary сonsiderably сalonе for is appropriatе i;;;Ы;hi.ь tь.y usе, and what anothеr. for .ui,to' may bе inappropriate

It is antiсipatеd that еvery readеr

;;i;;i;;;ЬiЫ ь"'

Thеrе arе many workеd ехamplеs in thе. Ьook, with сomplеtе, dеtailed answеrs to all the quеstions. At фе end ot еaсh worкro will find thе Ъymbol l to indiсate that thе сxam;;;;;ь: ple has beЪn сompletеd, and what follows is tеxt.

Somе

'.u of thе еxerсisеs from thе original Teaсh Yourself

i;й";";y ;;;й;"Ь

h"u. been usеd in this rJvised text.' but аll thе b..'' rеwoгked to takе aссount of the grеatеr

aссuraсy availaЬlе with сalсulators.

and I would likе to thank Linda Мoore fоr her hеlp in rеading

.o'i.сi''g

the tеxt. But thе rеsponsiЬility foг еrrors is mine.

Hugh Nеill

Novembеr 1997

сг

Цr

J

a 1+ o GI o o g) s)

II

о х

т

fl

-

C П 5 o.

ln this chаpter you will learn:

. .

what trigonometry is a Iittle about its origins.

1.2 What is trigonometry?

1.1 lntroduction

onе of thе еarliеst known examplеs-of thе P.raсtiсal 1PPliсa.tion ot onе ot fnе of gеomеtry was the problеm of finding thе hеtsht thе Grееk ThЪlеs, by Еsvotian Dyramros. iъi;;;; 'Ьlu"j (с.640 вс to 550 вс) using sim;f,ilБ;;;i'",',,"J-,tь.-atiсian ilar trianglеs.

Trigonomеtry is thе branсh of mathеmatiсs whiсh dеals with thе rеlationships Ьetwееn sides and angles of trianglеs.

For ехamplе, if yоu havе a right.anglеd trianglе with sidеs 7сm

and 5сm as shorpn in Figurе 1".2, trigonomеtry еnables you to сalсulatе thе angles of йе trianglе and thе lеngth of thе othет sidе.

figurc 1.2

You should Ье ablе to do thеsе сalсulations rмhеn you have studied Сhaptеr 02.

figure 1.1

pyramid and a Thales oЬsеrvеd thе lеngth of thе shadows of.thе еnd oI tnе thе at ground stiсk, АB, plaсеd vеrtiсally into фе 1.1'. Figure in shown shadЬw oi ih" py'"-id,

Similarl5 if you know thе information givеn in Figure 1.3 about the шianglе АBC, you сan сalсulate thе area of thе шiаngle, the lеngth of thе othеr side, and thе magnitudе of thе othеr angles. B

of thе shadow of thе pyramid BC rеprеsеnt thе lеngths .тБ. of thе pyr1щd is to thе height ;й;il;;йi. тh;i;'

oB

and

i.йь

the length of thе shadow ot thе pyramld of the stiсk, as '"iа

is tЬ thе lеngth of thе shadow of thе stiсk.'

That is, in Figurе 1.1,

PQ

BC AB=9!-.

known, you сan сalсulate PQ. at Thales' appli. !Ие arе told that thе king, Amasis, was аmazеd thе solution of to prinсiple abst,aсt oi .,,l"" lБЬmеtriсal As

QB, AB,

arld

BС

"', suсh a problеm.

are

idea of This idea is takеn up in Chapter 02, in introduсing thе the tangеnt ratio.

flgЦrc 1.3

You should bе ablе to do this whеn you havе studiеd Chaptеr 07.

1.3 The origins of trigonometry Aссording to George Ghеvеrghese Josеph in his book, TЙe Crest of the Peаcock, publishеd Ьy Pеnguin Books in 1990, thе origins of trigonomettУ arc obsсurе. A systematiс study of thе rеlationships Ьеtwееn the anglе at thе сеntre of a сiтсle and thе lеngth of сhord subtending it sеems to havе startеd rлrith Hipparсhus (c.750 вс). Sее Figurе 1.4.

+

сrd з6"

+

figure 1.4

Tablеs of thе lеngths of сhords for givеn anglеs wеrе produсеd Ьy Ptolеmy (с. ло 100). до Trisonomеtry bеgan to rеsemblе its prеsеnt form from about transwas knowlеdgе Thе t. is6 иtь thЁ wo-rk of Aryabhata mittеd to thе Arabs and thеnсе to Еuropе, whеrе. a сl€tarlес trtlе aссorrnt of trigonometriс knowlеdgе appеarеd undеr thе Rеgiomontanus. Ьy Бe triапgulЬ imni rnodis, writtеn io 1464

Orisinally trigonomеtry may havе Ьееn usеd to mеasurе thе ;;;;;';ъi,-Ё. f,.., of thе py'a*ids, but it was also usеd for astronomy.

Thе Indian mathеmatiсians Varahamihara (с..4D 500) and

I rlublishеd tablеs from whiсh it is possiblе to сomp.utе of anglеs. Thesе valuеs arе еxtremеly 1.:ч3::. Fo1 sinёs of u"fu., еxаmolе. thеir valuеs for sin 45o wеrе rеspесtlvеly U./v/U6 ano o.i oi to'' сompared with thе modеrn v a|trc, 0.7 07 |L.

A;йБй.;

i+ Цr J

o

1+

q) э

GT

o э

1+

. .

this chapter you will learn: what a tangent is the meanings of 'opposite',

.

'hypotenuse' in rightangled triangles how to solve problems

In

.adjaсent'and

using tangents.

2.1 lntroduсtion

pyramid in Thе method usеd by Thalеs to find thе hеight of thе today. usеd mеthod anсiеnt timеs is еssеntially thе samе as thJ is thе 2.1 Figurе сlosеly. more i;l;.h.;;i;'. rмorth .""йi''i,.g samе as Figure 1.1.

ffi

=

z.os.

PQ=2.05ХQB.

Thеrеforе

2'2 Тhe idea of the tangent rat.o The idеa of a сonstant ratio for evеry angle is thе kеy to thе developmеnt of trigonometry.

PoQ (Figure 2.2) Ьe any aсute anglе 0o. From points А, B, C on onе arm' say OQ, draw perpendiсulars АD, BЕ, СF t'o the othеr arm, OP. As thеsе pеrpеndiсulars arе parallel, thе triangles AoD, BoЕ and CoF arc similar. Lеt

Thеrеforе

АD

BE, CF oD-- oE -- oF'

figure 2.1

You сan assumе that the sunЪ rays arе parallеl Ьесausе the sun

it follows that thе islй;;;;-fio- th. еarth. In.Figw€ 2.tfаlling on thе tops of rays thБ ilj,iа ""6ipЫ*ьi.й ,.p,.'.''. thе objесts are parallеl.

АCB (thеy arе сorrеsponding thе altitudе of thе sun. reprеsent йсh ;;;й;j. fЁ"'еЪnglеs As anglеs PQB and ABC arc right anglеs, triangles PQB and АBC are similar, so

Thеreforе, angle PBQ = angle

'&=#"'#=#.

оf Thе height PQ of thе Pyrаmid.is indеpеndеnt of thе lеngth thе stiсk, the of АB ;й;;i;l.ъъ1f yoo Ёh,''g. the lеnglh

DЕF nguJe2.2

Now take any point У, it does not mattеr whiсh, on the arm oQ. For that anglе 0o the ratio of thе pеrpеndiсular ХY drarмn from Y on thе arm oQ to thе distanсе oХ interсeptеd on thе othеr arm

Бе сhangеd in propоrtion. .You сan important gеnеral dеduсtion. following thе makе

i.'*ii'_.r its shajow will

;ь;;;i;

For the givеn anglе АCB, thе ratio

ffistays

сonstant whatеvеr

AB. You сan сalсulatе this ratio Ьеforehand for nееd to usе thе stiсk, ;;;;йЪCB.]r й do this, you do notvaluе of thе ratio, and thе angfе,and ;;ъ#.i.;." r.".j* 'ье PQ. сalсulate you сan lengЬ the yoo йu.Ы.asurеd QB, for Thus if the аngle of еlеvation is 64. and the valuе of thе rаtio you havе then 2.05, Ье to found prеviously ь,jЪеen iьiТ,"gЁ

the lensth

of

figure 2.3

oP

is сonstant. Seе Figurе 2.3.

partiсular This is truе for any anglе; еaсh anglе 0. .Цч its own oг тnе tangеnt thе сallеd is ratio сorrеsponding to it. This ratio anglе 0".

In praсtiсе, thе namе tangent is abbrеviatеd to tan. Thus for 0" in Figurеs 2.2 aлd2.3 уotl сan writе

tan 0o =

2.з

YУ

бi.

of tangent

ratio Thеrе was a general disсussion of the idеa of thе tangеnt into disсussion to rеfinе.that -Ъ...i." z..i,ьtrt itof isthеimportant anglе. ot an tangеnt a formal dеfinition of radius In Figurе 2.4, the.origin o is thе се.ntrе of a сirсlе whеrе x-аxis, thе to 0o anglе an at 1 unit. Draw a ,,o,o,.oP y). (х' Ье P of сoordinatеs < thе 90. Lеt бJ o

^definition

As 0 inсrеasеs, y inсrеases and .r dесreasеs, so the tangеnts of angles сlosе to 90o are vеry large. You will sее that when 0 = 90, the valuе of л is 0, so } is not dеfinеd; it follows that

tan 90o does not ехist, and is undefinеd.

2.4 Yalues of the tangent You сan find фе value of thе tangent of an anglе by using youг сalсulator. Try using it. You should find that thе tangеnt of 45o, writtеn tan45o, is 1, and tan60o = I.732.... If you havе diffiсulty with this you should сonsult your сalсulator handЬоok, and make surе that you сan find thе tangеnt of any anglе quiсkly and еasily.

Your сalсulator must be in thе сorrесt modе. There arе othеr units, notably radians or rads, for mеasuring angle, and you must еnsurе that your сalсulator is in degreе modе, rather than radian or rad modе. Radians arе widеly usеd in сalсulus, and arе the subjесt of Chaptеr 05. Somе сalсulators also givе tangеnts for gradеs, anotЬеr unit for angle. Thегe are 100 grades in a right anglе; this book will not usе grades.

Your саlсulаtоr чrill also rеvеrsе this proсеss of finding thе tangеnt of an angle. If you nееd to know whiсh anglе has a tangеnt of 0.9, you look uP thе inverse tangent. This is often written аs tan-I 0.9, or sometimes as arсtan 0.9. Chесk that tan.10.9 = 41.987..... If it is not, сonsult your сalсulator handbook.

In thе work that follows, thе degree sign will always be inсludеd, Ьut you might wish to lеavе it out in your work, providеd thеrе is no ambiguiry Thus you wоuld urrite tan45o = 1 and tan60" = 1.732...

figurc2.4

by Thеn thе tangent of thе angle 0o, writtеn tan9", is dеfined

,^no =I. You сan sее from thе dеfinition that if 0 P is 0, so tanO = 0.

If0=45,thеnx=},So tan 45" = 1.

=

0 the y-сoordinatе оf

.

Еxerсise 2.1

ln questions 1 to 6, use your сa|сulator to find the va|ues of the tangents of the fol|owing ang|es. Give your answers сorreсt to three deсima| places.

I 3 5

tan 20" tan 89.99" tan 62o

2 4 6

tan 30" tan 40.43" tan 0.5'

questions 7 |o 12, use your сalсu|ator to find the ang|es with the tangents. Give your answer сorreсt to the nearest one

|n

tolЁwing

hundredth of a degree. 7

9 11

0.342 6.123 1

I

10 12

Thеn anglе

Thеn

PoQ

is thе anglе of еlevation and еquals 38.25.. Ь

fu=tan38.25. h

2 0.0001

= 168

lГз

2.5 Notation for angles and sides

It is Using notation suсh as АBC for an anglе is сumbеrsomе. the only u9i-ng anglg-lr an to rеfer to or*Ё й"'" сonvеniеnt ts no thеrе it it..Thus, definе whiсh three thе of lettеr middlе

,ЪЪis"i.y, tanB цrill bе usеd in prefеrеnсe for tanАBС. (theta) and ф Singlе Grееk lеttеrs suсh as с (alpha), B (Ьеta),0 (phi) are oftеn usеd for anglеs. h to Similarly, it is usually еasier to usе a singlе letter suсh as bеgingivе thе to than rathеr line, ."Ь'.'."i a distanсе alo',g a ning and еnd of thе linе as in thе form АB.

2.6 Using tangents

Hеrе arе somе еxamples whiсh illustra.tе the use of tangеnts ,"j iь" tесhnique of йlving problems with thеm. Exаmple 2.1 168m h"'i::"liJЦ^d':; A;;;.y* who is standing at a point measures the-angle of towеr aiall ;;;i';;.мi".i-Ъr r.--..:^rL^ tЬд inrrrрr яq ЗR 2-5o. hеisht the height Find ^J top ^f thе tower as 38.25o. "^. of ;й";i;;.ithе torлrеr. thе of яЬоve thе the ground sround of thе top abovе You should always draw a figurе. In Figurе.2.5, Pis thе tч o{ф9

= 168 x tan

=

.Гhе

38.25'

x 0.788зз64..-

1З2.44052,..

.

hеight of thе towеr is 132 m, сorrесt to thrее signifiсant figurеs.

l

ln praсtiсе, if you arе using a сalсulator, thеrе is no nееd to writе down all thе stеps givеn aЬovе. You should writе down сnough so that you сan follоw your own working, Ьut you do ll()t nееd to Writе down thе valuе of thе tangеnt as an intеrmе(liatе stеp. It is entirеly еnough, and aсtually bеttеr praсtiсе, to writе thе сalсulation abovе аs h

ffi h

=

tan 38'25"

= 168 x tan 38.25" =

1з2'44052...

.

llowеver' in this сhaptеr and thе nехt, thе еxtra linе will Ье

insеrtеd as a hеlp to thе rеadеr.

Ехamp|e2.2 A pеrson who is 168сm tall had a shadow whiсh lt>ng.

Find the anglе of еlеvation of thе sun.

P

';й

168 сm

ligure 2.6

ligшъ 2.5

154сm

ln Figurе 2.6 |et PQ Ье the pеrson and oQ bе thе shadorм. Thеn /,o is thе sunЪ ray and 0o is thе anglе of еlеvation of thе sun.

wluсп towеr and Q is thе boшom. Thе survеyоr ls standrng at L,, h metrеs. be towеr thе of hеight йе Let as levеl ;;' th" Q.

168 m

rлras

Then

1,68

Thеrеforе:

tan0" =т54 = 1.09090...

0

47.489...

1a11Q'=$ - 154 o

=и"'(ffi) =

=0.624869... I

.

Thеreforе thе anglе of еlеvation of thе sun is approximatеly 47.49".1 out a Notе onсe again that you с1n usе thr сalсulator and lеavе to еxplanation еnough. you that givе p'ouid"d siеps, of ,'oйb.' writе сould you ;Б;; h.* yori obiain your rеsuit. Thus

47.489...

Example 2.3 Fisurе 2.7 reprеsеnts a seсtion of a symmetriсal roof in whiсh АЕ;h; 'p",''' "''d oP thе risе. Р is tЬe mid-point of АB.

Thеriseоfthеroofis7manditsanglеofslopeis32..Findthе roof span.

=2,

*=#ь

so

= tan-l1.09090...

=

tan32o

= 11.2023...

.

Thе roof span is 2ш metres, that is approximate|у

22.4m.I

Еxerсise 2.2

1 2

The angle of elevation of the sun is 48.4". Find the height of a

flag staff whose shadow is 7.42m long. A boat leaving a harbour travels 4 miles east and 5 miles north. Find the bearing of the boat from the harbour.

3 A boat whiсh is on a bearing of 038" from a harbour is 6 miles north of the harbour. How far east is the boat from the harbour? 4 A ladder resting against a wall makes an angle of 69' with the ground. Тhe foot of the ladder is 7.5 m from the wal|. Find the height of the top of the ladder. From the top window of a house whiсh is 1.5 km awаy from a tower it is observed that the angle of elevation of the top of the tower is 3.6o and the angle of depression of the bottom is 1.2". Find the height of the tower in metres.

From the top of a с|iff 32 m high it is noted that the angles of depression of two boats |ying in the line due east of the с|iff are 21'and 17'. How far are the boats aparl? Тwo adjaсent sides of a reсtang|e are 15.8 сm and 1 1 .9 сm. Find the ang|es whiсh a diagonal of the reоtangle makes with the sides. P and Q are two points direсt|y opposite to one another on the banks of a river. A distanсe of 80 m is measured a|ong one bank

figшв2.7

at right angles to PQ. From the end of this line the angle

OP As thе roof is symmetriсal'oAч is an isosсеles trianglе, so metrеs. tll АP, lеngth is pеrpеndiсulai to АB. Call thе

A |adder whiсh is |eaning against a wall makes an ang|e of 70"

subtended by PQ is 61'. Find the width of the river.

with the ground and reaсhes 5 m up the wal|. The foot of the |adder is then moved 50сm c|oser to the wa|l. Find the new angle that the ladder makes with the ground.

2.7 opposite and adiaсent sides

work is not сonSometimеs thе trianglе with whiсh yоu havе.to an еxample of shows p"g..гig"'.2.8a

;j;;,l'y;й"й"i.i. this.

А

/l

/

ang|е 0f focus

1,.'

Ь1

tanBo=oЧPo'i..=9. - adlaсеnt )

tigurе 2.8b

In this сase, thеrе is no сonvеniеnt pair of axеs involved. il.;;;.;;;;.Ьold 'o,,t" the figurе, еi1hеr aсtually or in your tо oЬtain Figurе 2.8b. imagination,

p,

сalсulate = I.6,and you сan You сan now sее thattanB' =|5o"-E thе diafrоm еasily _'Lцдr,. = Lrr4L t^iB" seе that сould you how LUцru but Duf, ll()w ).vu D9U !s = r.с

prосеss of gram in Figurе 2.8a,,without going tйrough thе !еtting to Figurе 2.8Ь? alwаys.bе \X/hеn you arе using a right-anglеd triangle уou.will tоr angtе. rrght thе than othеr anglеs thе of intеrеsiеd in onе sidеs thе of of.foсrrs'. onе ;;;;;;;;, .,ir ,ьь "',gl. Тh. jзчc!. sidе thе oppositе. onе of .all tйs will Ье oppositе tьi' ""?1Ъ; yoц arе lntеrеstеq; whiсh in join anglе thе thе othеisidеs will сall this sidе thе adiaсеnt.

Thеn

This works for

tangеnt

-

=

figure 2.9b

As you сan ser' in both сasеs

5сm

figure 2.84

adjacent

flgшrc 2.9a

Мany pеoplе find this method thе most сonvеniеnt whеn using

thе tаngent.

.Гhе

other side of thе right.anglеd trianglе, thе lоngеst sidе, is

сallеd thе hдlotеnuse. The hypotеnusе will fеaturе in Chaptеr 03.

Еxample 2.4 ln a triangle АBC, angle B = 90",АB = 5сm and BC = 7 cm. i.ind thе sizе of angle А. l)raw a diagram, Figurе 2.10.

5сm B

opposrfе

-=_-_-:. adlaсеnt

all right-anglеd trianglеs. In thе two сasеs in 2.8:й thе"оppositе-and adiaсеnt sidеs arе

F;'; z.в, ""а laБеllеd in Figurе 2.9a and 2.9Ь.

|lщm2.l0

ln triangle АBC, foсus on anglе А. Thе opposite is 7сm and thе adiaссnt is 5сm. 'Гhеreforе irnd

,^nА" anglе

=Z5

A = 54.46.I

Note that in this сase you сould find angle C first using tanC" -- |, АBC and thеn use thе faсt that thе sum of thе anglеs of trianglе А. angle is ].80. to find

tan52o _ 13.3

Thеn

**

_ -

'с

1,3.3

tan 52" 13.3

Example 2.5 Find thе length а inЕiglхe 2.It.

1.2799... = 10.391...

.

Thеrеfore the lеngth is 10.4сm, сorrесt to 3 signifiсant figurеs.

Еxerсise 2.3 |n

figurc 2.11

----1"' scm

Foсus on thе ang|e 24". Thе oppоsitе sidе is a and thе adjaсеnt sidе is 6 сm.

tan24"

Then

a

= Ч

3

6

=

6 tan24"

= 6х0.4452...

Therеforе а

=

=2.671.... . 2.67 cm,сorrесt to 3 signifiсant figures.

l

Еxample 2.6 Find thе lеngth xсm in Figlte 2.L2.

eaсh of the folIowing questions find the marked ang|e or side.

6

'

П-__qJ_9щ--.-?- 4

hсmt"""

---.

*,fuf /сm

Lт---__-_-

Foсus оn the angle 52". The oppositе sidе is 13.3 сm and thе adjaсеnt sidе is

rсm.

# т

4.8 сm

*.. 4.2cm.---7 u,"

6

Uю\

ro figurэ 2.12

,r,'1or

lro

l" \

д^tt - \xсm 2сm/

6сn/ N сm д.\ ,/

or'/f

et //l

t

3.1 lntroduсtion In Figurе 3.1 a pеrpendiсular is drawn from

You

sarдr

on pagе 8 that the ,^tio

ffi=

А to OB.

tan0o.

Now сonsidеr thе ratiоs of еaсh of thе lines АB and oB to thе

ОA

hypotеnuse

trianglе

of'

oАB.

o

B

flgure 3.1

oa o ao g) o

'|ust as for a fixеd anglе 0o thе ratio tan 0") wherеvеr

Пl

А is,

so also thе

is сonstant.

,Гhis

э

4Д i,.o,,,1"nt, 'i{i"

p,that

\JI\

is

(and еqual to

nypotеnusе

ratio is сallеd thе sinе of thе anglе 0o and is written sinOo.

f

э

a

э

a.

sin0o = tlgure 3.2

|n

. .

.

this сhаpteryou wil| leam: what the sine and сosine are how to use the sine and cosine to find lengths and angles in right-angled

triangles

how to solve multistage problems using sines'

сosines and tangents'

a

opposite hypotenuse

-

Similarly, thе ratio

=

с0s0o =

Б

Д

adiacent =

hypotenuse b

.\B adiaсent !{, г::t::::. ' is also сonstant |JI\that is nypoЕеnuse

for

thе angle 0o. 'Гhis ratiо is сallеd thе сosinе of the angle 0o and is wriшеn сos 0o.

.Гhus

sin9o =

Jpp'osце

nypofеnrrsе=

g, D

сos9o = Дn,,"t =g. hypotеnusе b

3.2 Definition of sine and сosine

о

[n Sесtion 3.1 thеrе is a short disсussion introduсing thе sinе and сosinе ratios. In this sесtion thеre is a more formal dеfinition.

Dгaw a сirсlе with radius 1 unit, and сеntrе at thе origin O. Draw thе radius OP at an anglе 0o to the x-аxis in ai antiсloсkwise dirесtion. Sеe Figure 3.3.

Using PythagorasЪ йеorеm on trianglе

Thеrеfore sin2Oo

oPN

givеs

* +f

= 1.

сos20o = 1, where sin2 0" mеans (sin 0.)2 and means сos2 0o mеans (сos0.)2.

Thе еquation thеorеm.

sin2

0"

+

+ сos2 0o = 1'

is oftеn сallеd Pythagoras's

Finding thе valuеs of the sinе and сosinе of anglеs is similar to finding thе tangеnt of an anglе. Use your сalсulator in thе way that you would еxpесt. You саn ujе the funсtions sin-1 and сos-l to find thе invеrsе sinе and сosinе in thе same wav that you used tan-1 to find thе inversе tangent.

3.3 Using the sine and cosine In the еxamplеs whiсh follovy therе is a сonsistent stratеgy for starting thе problеm.

о Look

at thе angle (othеr than thе right angle) involvеd in the proЬlеm.

о

Idеntify. the. sidеs, аdjaсent, oppositе and hypotenusе, involvеd in thе problеm.

о

Deсidе whiсh trigonomеtriс ratio is detеrmined by thе rwo

figure 3.3

Thеn lеt P havе сoordinates (r, y).

Then sin 0. = y and сos 0o = х ate thе dеfinitions of sinе and сosinе whiсh will Ье used in thе remaindеr of thе book.

Notе the arrow labelling thе anglе 0. in Figure 3.3; this is to еmphasize that angles are measured positivеф in the antiсloсk-

rмisе direсtion.

Notе also fwo other properties of sin0o and сosOo.

о

In thе triangle

sin(90

oPN, anglе oРN

-

0)o

= (90

- 0)., and

сos0o, ='-Д.$ nypoЕenuse= 9l =

that is

sin(90_0)o=сos0o.

Similarly

сos(90_0)o=sin0o.

sidеs.

о Мakе.аn

еquation rмhiсh starts with thе trigonomеtriс ratio for the angle сonсerned, and finishеs with thе division of two lеngths.

о

Solve the еquation to find what you neеd.

Hеrе arе somе examples whiсh usе this stratrgy.

Exаmple 3.1

Find thе 3.4.

figurc3.4

lеnф markеd ясm in

the right.anglеd шiangle in Figurе

Thе anglе сonсеrnеd is 51"; rеlativе to thе angle of 51o, thе sidе 2..5 сrrr is thе adjaсеnt, and thе sidе markеd x сm is thе ltyp
сos51" =2'5 х

.

Example 3.3

A 30m laddеr on a firе engine has to rеaсh a.V/hat window 26m from thе ground whiсh is horizontal and lеvеl.

anglе, to

thе nеarеst degreе, must it makе with the ground and how far from thе Ьuilding must it Ье plaсеd?

AP (Figurе 3.6), lеt 0. bе thе anglе that thе laddеr makеs with thе ground and lеt d mеtrеs bе the distanсе of thе foot of thе laddеr from thе window. Lеt thе laddеr be

Thеn solvе this еquation fоr x.

* = -2!L- = 3'972"' сos.)l-

'

Thе lеngth of thе sidе is 3.97сm аpproхimatеly.

l

Еxample 3.2 Thе lеngth оf еaсh lеg of a stеp laddеr is 2.5 m. !Иhеn thе lеgs arе opеnеd out' thе distanсе Ьеrwееn thеir fееt is 2 m. Find thе anglе bеrwееn thе lеgs.

In Figurе 3.5, lеt

АB

and

АС

Ье thе lеgs of thе laddеr. As thеrе

Аdm

is no right anglе, you havе to makе onе Ьy dropping thе pеr. pendiсular Аo from А to thе Ьase BC. Thе trianglе АBC is isosсеlеs, so Аo Ьisесts thе anglе BAС and thе basе BC.

figurв 3.6

BО=ОС=1m.

Тhеn so

Thеrеforе

А

As thе sidеs 25m and 30m arе thе oppositе and the hypotеnusе for thе anglе 0o, you neеd the sinе ratio.

.Гhе "Гo

ang|e

hypotеnusе for thе ang|e do' so you nееd thе sinе ratio.

Thеn

so and

JU

0" = 60.07З...

.

find thе distanсе d, it is best to usе PythagorasЪ theorеm.

*

= 302

-

262 = 900

-

676

=

244,

d=14.966....

Thе foot оf the laddеt is 14.97 m from thе wall.

BАС. Сall it 2d", so ang|e BAО = df, Thе sidеs of lеngth 1 m and 2.5 m arе thе oppositе and thе You nееd to find

Ц=0.8666.,.

ladder is plaсеd at an anglе of 60" to thе ground.

so

figure 3.5

sin0o=

sin do

=

а, =

*=

Еxample 3.4

Thе hеight of a соnе is Find thе slant hеight.

O.*

23.578...

,

2a=47.156....

Thеrеforе thе anglе bеtwееn thе lеgs is 47.16. approхimatеly. I

figun 3.7

1.8

l

сm, and thе anglе at thе vеrtеx is 88".

In Figure 3.7,|et /сm bе thе slant hеight of thе сonе. Sinсе the pеrpёndiсular to the basе Ьisесts the vеrtiсal angle of thе сone, еaсh part is 44".

Thе sides 18сm and lcm are the adjaсеnt and hypotenusе for the angle 44o, so thе ratio сonсеrned is thе сosinе.

Thеn

сos44o

/=***.4.

=

+,

Thе slant hеight is approximatеly 25.0сm.

1т

l

18 A person is walking up а road angled at 8" to the horizontal. How far must the person walk along the road to rise a height of

Exerсise 3.1

ln questions 1 to 10, find the side or angle indiоated by the |etter.

bу
o,Ч,41'

,t

r------7

4 4'*,'V^K

{-_---7

6

13сm

N'1o1

.-.\\ 11 A

\,/z'lз',.

,Ьt

/*.

\xсm

4.6

m and 5.8 m. Find the аngles and the |ength of the

hypotenuse.

3.4 Тrigonometriс ratios of 45o, 30" and 60"

You сan сalсulatе thе trigonomеtriс ratiоs еxaсtly for somе simplе anglеs.

Sinen Gosine and tаngent of 45" Figurе 3.8 shows an isosсеles right-anglеd triangle whose еqual

sidеs are eaсh 1 unit.

1зсm

{,,^

'.^v

сirс|e of radius 45mm has a сhord of |ength 60mm. Find the

sine and the сosine of thе ang|e at the centre of the сirс|e subtended by this сhord.

1 km?

19 ln a right-ang|ed triang|e the sides сontaining the right ang|e are

Jo"'

/ ' I:у<;; I ,' \ ,.d\ "f--^ /сm

14

16

I =25.022....

7---7

13

15

=т7&:

2cm "

|n a сirсle with radius 4 сm, a оhord is drawn subtending an

angIe of 80. at the оentre. Find the length of this сhord and its distance from the сentre. Тhe sides of a triangle are 135 mm, 180 mm and 225 mm. Prove that the triangle is right-angled, and find its angles. ln a right-angled triang|e' the hypotenuse has |ength 7.4сm' and one of the other sides has length 4.6cm. Find the smallest angle of the triangle. A boat travels a distance of 14.2km on a bearing of 041'. How far east has it travelled? Тhe height of an isosсeles triangle is 3.8 cm, and the equa| angles are 52". Find the length of the equal sides. A сhord of a cirсle is 3 m |ong, and it subtends an angle of 63. at the сentre of the оirоle. Find the radius of the сirс|e.

figurв 3.8

Using PрhagorasЪ thеorеm, thе hypotеnuse has lеngth {2 units. Thеrеforе thе trigonomеtriс ratios fоr 45. arе given by sin45o

=

1},

сos45"

=#,иn45o

Whеn you usе a сalсulator to solvе an еquation suсh as sin 0" = is important to Ьe ablе to gеt аs aссurate afl answеr as #,'. you сan.

= ]..

You сan, if you wish, usе the equivalеnt valuеs sin 45o

=$,,o,45" =Ч,tan

3.5 Using the сalсulator acсurateIy

Wrong method

45o = 1.

Thesе valuеs arе oЬtainеd from thеir prеvious valuеs by noting

that

1.

a,=a,' 1,

lт

12

=

fl

=

o.вzz

sin0o = З8.46.

Gorreсt method

E=т'

sin,o

sin,o

=

ff

=

o.вztв27621,6

0 = З8.43.

Sine, сosine and tangent of 30" and 60" Figurе 3.9 sholмs an еquilatеral trianglе of side 2 units.

\07hat has happеnеd? Thе proЬlеm is that

in the wrong method thе сorrесtеd answеь 0.622, a thrеe-signifiсant-figurе approxi-

mation, has bееn usеd in thе sесond part of the сalсulation to find thе anglе, and has introduсеd an еrror. Yoц сan avoid the еrror Ьy not writing dоrмn thе thrеe-signifiсantfigure approximation аnd by using thе сalсulator in thе follow. lng way.

sinl =ff

0 = 38.43.

'I

1

In this vеrsion' thе answе

figure 3.9

' '" {was

usеd dirесtly to сalсulate thе

anglе, and thеrеfore all the figйiеs wеrе prеsеrvеd in thе proсess.

Thе perpendiсular from thе vеrtеx bisесts the basе, dividing thе origiЬal.trianglе into two trianglеs with anglеs of 30o, 60" and 90; and sidеs of lеngth 1. unit, 2 units and, using Pythagoras's theorеm, V3 units. Thеrеfore thе trigonomеtriс ratios for 30o are givеn by

сos30. sin3Oo = },

=$,tan30.

=

тЬ

"'f

.

Thе samе ratios for 60" arc givеn by sin 60o =

f,

сos 60 " =

\,

tan60o = {З

.

It is usеful еithеr tо rеmеmber thеsе rеsults' or to be ablе to get thеm quiсkly.

Somеtimеs it may bе neсеssаry to usе a сalсulator memory to storе an intеrmеdiatе answеr to as many figures as you nееd.

tt is not nесеssary in this саsе, Ьut you сould сalсulate

ff

put the rеsult into mеmory А. Thеn you сan сalсulatе sin-1

нсr an aссuratе answеr.

a"d

А to

Y
3.6 Slope and gradient l;igurе 3.10 rеprеsеnts a side viеw of thе seсtiоn of a rising path AB is horizontal and BС is thе vеrtiсal risе.

^С'

f,guш 3.10

As

aor0t

so

АC ЕF

Thеn 0o is сallеd thе anglе of slopе or simply thе slоpе оf thе path.

Thе ratio tan 0o is сallеd thе gradient of thе path.

.!Иhеn

In

oсm

4.B сm

| --'pp-/псn

takе sin 0" instеad of tan 0" as thе gradiеnt.

sin0. (Ьy mеasuring BC and AС), and thе differеnсе Ьеtwееn AС and АB is rеlativеly

In praсtiсе also it is еasiеr to mеasurе

4.2cml -,.

IY

д\ хcm/ \ 160'

ГI------ /=3.52,^|вв'l"'1

small provided that 0" is small.

You сan use your сalсulator to see just how small thе diffеrеnсе is Ьеfwееn sinеs and tangеnts for small anglеs.

4.2cm

1)r

уcn/

/5o.l'

3.т Projeсtions AB

4сm

5.1 cm

way and most roads' it makеs littlе praсtiсal diffеrеnсе if you

/

,/

\

,о.t

.lз

\

Хсm p.\ cm

10

Ьe a straight

ln a right.angled triangle the two sides сontaining the right angle have lengths 2.34m and 1.64m. Find the smallest angle and the length of the hypotenuse. 12 |n the triangIe ABс, c is a right ang|e' АC is 122сm and АB is 11

ffI figure 3.11

Pеrpеndiсulars arе drawn ftom А and B to /, meeting l at F. Thеn ЕF is сallеd thе projeсtion of АB on /. еqual.

2

4оm

the anglе of slopе 0" is very small, as in thе сasе of a rai|-

You сan sеe from Figurе 3.11. tЬat thе lеngths

questions 1 to 10, find the marked angle or side.

--/-1,.rr'

+.

In Figurе 3.7\,|et / bе a straight linе, and lеt linе sеgmеnt whiсh makеs an anglе 0. with /.

АB cos9" = AB сosOo. =

tangents.

Somеtimеs, еspесially by thе side of railways, thе gradiеnt is givеn in thе form 1 in 55. This mеans =

ff,

Еxercise 3.2 Тhis is a misсe|laneous exerсise involving sines, сosines and

l-еt thе anglе Ьеtwеen thе path and thе horizontal bе 0o.

sradiеnt *55

=

АC

aпd

Е and

ЕF arc

175сm. Calсu|ate ang|e B. АBC' angle C = 90o, zlnd А = 37.35. and AB = 91 .4 mm. Find the |engths of BC and CА. '14 А8C is a triang|e. Angle C is a right ang|e, АC is 21.32 m and BC is 12.56m. Find the ang|esА and 8.

13 ln triangle

15 ln a triang|e ABc, АD is the perpendiсu|ar from А to BC. Тhe longths of АЕl and BC are з.25 сm and 4.68 сm and angle B is 55'. Find the lengths ot AD, BD and AC. 16 АBC is a triang|e, right.angled at C. The |engths of BC and AB are 378 mm and 543 mm' Find ang|e А and the |ength of GА. 17 A |adder 20 m |ong rests against a vertiсal wa|l. Find the inсlination of the |adder to the horizonta| when the foot of the ladder is 7m from the wall. 18 A ship starting from o travels 18 km hj in the direоtion 35. north of east. How far will it be north and east of O after an hour? 19 A pendulum of length 20cm swings on either side of the vertiсal through an angle of 15". Through what height does the bob rise? 20 Тhe side of an equi|ateral triangle is x metres' Find in terms of x the altitude of the triang|e. Henсe find sin 60" and sin 30o. 21 A straight Iine 3.5 сm |ong makes an angle of 42" with the x-axis. Find the |engths of its projeоtions on the x- and y-axes. 22 When you walk 1.5 km up the line of greatest slope of a hill you rise 94m. Find the gradient of the hill. 23 A ship starts from a given point and sails 15.5km on a bearing of 319.. How far has it gone west and north respeсtive|y? 24 A point P is 14.5 km north of Q and Q is g km west of Е. Find the bearing of P from R and its distance from R.

сс flguтe 3.12

The lеft-hand diagram in Figurе 3.12 sholмs thе information. In thе right-hand diagram, the diagonals, whiсh сut at right-angles at O, havе bееn drawn, and the line АC, whiсh is an aхis of symmetry' Ьiseсts thе kitе. Lеt

Do

bе

rсm, and lеt

so

This sесtion givеs examplеs of multistagс proЬlеms whеrе you neеd to think out a strategy bеfore you start.

Also

You nееd also to bе aware of the adviсе given in Sесtion 3.5

Substituting for

Example 3.5 АBCD is a kite inwhiсh AB = АD = 5сmandBC Anglе DАB = 80o. Calсulatе anglе BCD.

x

СD =7сm.

Angle BCD

=

sinO'=

so =

DCA

sin40'=

Thеn

3.8 Multistage problems aЬout thе aссuratе usе of a сalсulator.

ang|e

x

sinoo

=

0".

! 5 sin4Oo.

f

.

=:rЭ40.

0=27.З3.... =

20o = 54.66o, сorrесt to two dесimal plaсеs.

l

Example 3.6 }.igurе 3.13 rеpresеnts Part of a symmеtriсal roof frame. PА = 28 m, АB = 6m and ang|e oPА = 21". Find thе lengths of oР and oA.

Еxerсise 3.3

1

2

ABOD is a kite in whiсh AB = AD = 5 сm, BC = CD = 7оm and ang|e DAB = 80". Ca|culate the length of the diagonalАC. ln the diagram, find the |ength of

АC.

t

4сm figurc 3.13

Lеt OP= lm, PB

=

oА

trm xnd

=

I

hm.

To find / you neеd to find ang|e oPB1' to do this you nееd first to find ang|e АPB. Lеt anglе АPB= d".

Thеn so

3

ln the diagram, find the angle 0.

sindo= L=0.21,428... JO

d=1.2.373....

Thеrеfoге

anglе

oPB

=

d"

=

33.373...'.

+ 21"o

PQЛS is a reсtangle. A semiсirсle drawn with PQ as diameter .t0сm' сuts яs at А and B. Тhe |ength PQ is and angle 8QP is 30.. Calculate the |еngth PS.

Nеxt you must find thе lеngth.r.

A ship sai|s 5 km on a bвaring of 045" and then

To find x, usе PрhagorasЪ thеorem in trianglе АPB. so

х

Thеn

сos33.373..

=

27.349..,

point.

,

I

.|0cm.

l= 27.349... сos 33.373.... = З2.7500,.. .

b,уou nееd to

To find OB,

=

oB

_ 6.

=

т#o^

OB

=

32.7590... x sin 33.373..."

=

19.0156...

,

h = 18.0156.".

-

and

oP

oB - AB

=

Calсu]atethe length of АC.

thо isosсeles triang|e АBC, the щual angIes at B and C are eaсh 50o. The sides of the triangle each touоh a сirоle of radius 2 оm. In

sin33.373....

so

Thеrеfore

usе Й =

on a

Тhe |engths AB and AC oI a triangle ABC are 5сm and 6сm respeсtive|y. The |ength of the perpendiсu|ar from А to BC is 4сm. Calсulate the angle BАC. ln a triangle АBC, the angles at А and C are 2o" and 30. respeсtive|y. Тhe length of the perpendiсu|аr from B to АC is

.' - 27.3!9...

so

6 km

bearing of 060.. Find its distanсe and bearing from its starting

x2=282-62=784-36=748

To find

B*8сm.+

6 = 12.0156...

32.75^ and oA = 12.02m approхimatеly.

l

Calсu|ate the |ength BC. i

A ladder of length

5 m is |eaning against a vertiсa| wal| at an angle of 60' to the horizontal. The foot of the Iadder moves in by

50сm. By how muсh does the top of the |adder move up the wall?

PQЕS is a trapezium, with PQ parallelto RS. The angles at P .|30. respeсtively. Тhe length PQ is 6.23cm and Q are 120. and and the distanсe between the parallel sides is 4.92 cm' Ca|сu|ate the length of ЕS.

o. э fl

fl

1+ Цr J

o o a o rl o a з

э

э

-

In

.

. .

this сhapteryou will learn: the impoftanоe of good diagrams in solving threedimensional problems how to break down threedimensional problems into

two-dimensional problems how to solve problems by using pyramids, boxes and wedges.

4.1 lntroduction

Thеrе arе a numbеr of fеaturеs you should notiсе about this

\DИorking in thrее dimеnsions introduсеs no new trigonomеtriс

diagram.

idеas, Ьut you do nееd to bе ablе to think in thrее dimеnsions and to Ье aЬlе to visualizе thе proЬlеm сlеarly. To do this it is a grеat hеlp to Ье aЬlе to draw a good figurе.

You сan solvе all thе proЬlеms in this сhaptеr by piсking out right-angled trianglеs from a thrеe-dimеnsional figurе, drawn, of сoursе, in two dimеnsions. This сhaptеr will сonsist mainly of ехamplеs, whiсh will inсlude сеrtain typеs of diagram whiсh you shоuld Ьe aЬlе to draw

quiсkly and еasily.

о

Thе diagram is largе еnough to avoid points bеing .on top of onе anothеr'.

о

Thе vеrtеx V shоuld bе abovе thе сеntrе of thе Ьasе. It is bеst to draw thе basе first, draw thе diagonals intersесting at O, and thеn put V vertiсally abovе O, siting V so that thе edgе VС ot thе pyramid does not appеar to Pass through B.

о

Note that thе dimеnsions havе not bееn put on thе mеasufеmеnts whеrе, to do so, would add сluшеr.

To solvе thе problеm' You must dеvеlop a stratеgy whiсh

4.2 Pyramid problems

i

Thе first type of diagram is the pyramid diagram. This diagram

will work for all problеms whiсh involvе pyramids with

a

squarе Ьase.

nvolvеs сrеating or rесognizing right-angled trianglеs.

Thе anglе АVC is thе vеrtiсal anglе of thе isosсеlеs triang|e АVС whiсh is bisесtеd Ьу VО.If you сan find thе lеngth Аo, you сan find ang|e AVС. But you сan find Аo by using the faсt tЬat ABСD is a squarе. Sее Figurе 4.2.

Еxаmple 4.1

ABCD is thе square

base of sidе 5 сm of a pyramid whosе сm direсtly abovе thе сеntrе o of thе squarе. Calсulatе thе angle АVC.

Yertex V is

,/"1\'-lxl ,Ьtt

6

Thе diagram is drawn in Figurе 4.1.

А

g

с

А-5сm-в

ligure 4.2

lrind

АC by using Pythagoras's thеorеm

Аc=52+52=50,

so l.еt anglе

АC

АVo

Тhеn and Thеrefore figure 4.1

=

=

{50сm

d".

and'

Аo

=

}r/50сm.

|

t^nа"

=i\)0 b

o=30.5089....

2do = 6L.02".. If you find that you сan sоlvе thе problеm without drawing thе subsidiary diagram in Figurе 4.2, then do so. But many people find that it hеlps to sее what is happеning. ang|e

AVС

=

All

squarе-bаsеd pyramid problems сan be solved using this diаgram.

Example 4.2 РQRI'V is a pyramid with vеrtеx V whiсh is situatеd symmеtri сally aЬovе thе mid-point o of thе rесtangular Ьasе PQRS. Thе lеngths of PS and RS arе 6 сm and 5 сm, and thе slant hеight VP is 8 сm. Find thе anglе that thе еdgе VP makеs with the ground. Figurе 4.3 shows thе situation.

As

Po = ,PR,

and, using PщhagorasЪ thеorеm gives

you find

Thеn so

PR

=

{FУ;тsz =la +т

Po

=

+{6псm.

сosсo q,

=P;i

=^[61.

=+,

= 60.78.

Thеrefore thе еdgе VP makеs an angle of 60,78" with thе zontal. I

hori

4.3 Box problems The seсond typе of problem involvеs drawing a box.

figure 4.3

Let the rеquirеd anglе VPo bе d". Thеn you сan find d from triangle vPo if you сan find еithеr PO or Vo. You сan find

flgшв.|.5

PO from thе reсtangle РQRs.

Thе usuаl way to draw thе box is to draw two idеntiсal rесtan. gles, onе .bеhind' the othеr, and thеn to join up thе сornеrs appropriatеly. It сan bе usеful to makе sоmе оf rhe lines dottеd to makе it сlеar whiсh faсе is in front.

,^ tigure 4.4

|ж:

Еxаmple 4.3

A roоm has length

4 m,

width

3

m and hеight 2 m. Find thе

anglе that a diagonal of thе room makеs with the floor.

This figurе is part of thе bох diagram in Figure 4.5, with somе linеs rеmoved. F is thе foot of thе pylon. Tо сalсulatе SW, start by finding thе hеight of thе pylon, thеn thе lеngth FW, and thеn usе Pрhagoras's thеoтеm to find SW.

А

In trianglе

PF$

In trianglе

PFIV

4m

figure 4.6

Lеt thе diagonal of thе room bе АR, and lеt AС Ьe thе diagonal of thе floor. Lеt thе rеquirеd angle Ьe сo.

You сan usе Pрhagoras's thеorеm to find thе diagonal thе floor and so find thе anglе 0".

AC

=lT +4

tand" =Z

and

d

=

21.80.

Thеrеforе thе diagonal makеs an anglе of 2]".80. with thе floоr.

=

i#=ryщ

=

1.0o.

50tan10"

=

FW +sF,

=

(w),

= 12654.96,

Sw

= 172m. сorreсt to the nеarеst mеtrе.

*'o,

l

4.4 Wedge problems l

Somеtimеs the problеm is a Ьox proЬlеm, Ьut may not sound

likе onе.

Thе third typе оf problеm involvеs drawing а wеdgе. This wеdge is rеally only part оf a box, so you сould think of a wedgе pтoblem as a speсial сasе of a box problеm, but it is еаsiеr to think of it in a sеparatе сatеgofy.

ЕxampIe 4.4 The top, P, of a pylon standing оn lеvеl ground suЬtends an angle of 10o at a point S whiсh is 50m duе south, and 5o at a pоint W lying wеst of thе pylon. Calсulatе thе distanсе STV'сorrесt to thе nеarеst mеtrе.

flgшв4J

figure 4.7'

tanso

sР

Finally

so

ff

FW =---. tan 5-

АC of

=^'[I5 = 5

50 tan

=

тfr

so

,o PF

tan 1.0o

ttample 4.5 The line of greatest slopе of a flat hillside slopes at an anglе of 20o to thе horizontal. To rёduсe the anglе of сlimb, a walkеr walks on a Path on the hillside whiсh makes an angle of 50o with thе line of grеatest slope. At what angle to the horizontal doеs the walkеr сlimb on this path?

Ес

Aс= In triаngle

so

АBC

Бzб*mi.

sinс =;fu а

=

=

sin 20o сos 50o

=

sin20oсos50"

|2.70.|

Thе walkеr walks at |2.7" approximatеly to фе linе of grеatеst slopе.

You may fiпd it еasiеr to follow if you еvaluatе АE and АC as you go along, but it is bеttеr praсtiсе tо avоid it if you саn.

Exerсise 4 1 A pyramid has jts vertex direсtly above the сontre of its square base. The edges of the base are вaсh 8сm, and the vertiсal

fiцп{9

2

I-еt АЕ bе thе linе of greatest slope, and АC bе thе path of the walker. Thе angle that you neеd to find is anфe BAС. Call this angle do. As there are no units to the problеm, let the hеight BC bе 1 unit.

3

d you neеd to find anothеr length (apart from BC) in trianglе BАC. This п,ill сome first from thе rфt-angled trianglе DAE, and then from tгianglе AЕC, shown in Figure 4.10. Note that angle DAE = 20o as АЕ is a line of greatеst slope.

4

To find

,

ngп:з410

In triangle In triangle

АЕD ЕАC

6

hвight is 10сm. Find the ang|e between the s|аnt faсe and the base, and the anglo between a slant edg6 and the base. A symmetriсal pyramid stands on a square base ol side 8 сm. Тhe sIant height of the pyramid is 20cm. Find the angle between the s]ant edge and the base, and the angIe between a s]ant faсe аnd the base. A square board is suspended horizontally by four equаl ropes attaсhed to a point P direсtly above the оentre of thв board. Еaсh rope has length 15m and Ь inсlined at an ang|e of 10" with the vertiоal. Calсulate the tengith of the side of the square board. A pyrаmid has its vertex direсtly above the сentreo' its square base' Тhe edges of the base аre eaсh 6 оm, and the vertiсa] height is 8оm. Find the angle between two adjасent slant faсes. Find the angle that a main diagonal of a сube makes with the base. 1д56u'e that the сube has sides of length 1 unit.)

0A

;,,-J^7' sin2Oo =

11

7i

so

АЕ 7

сos50o =

=

6fuа.

/r'W=;йт?, AC

".^2

pylon is situated at a сornar of a reсtangular fie|d with

dimensions 100m by 80m. The angle subtended by the pylon at the furthest сorner of the field is 10". Find the angles subtended by the pylon at the other two сorners of the field, т A regulаr tetrahedron has a|} its edges 8 оm in length. Find the angIes whiсh аn edgв makes with the bаse. 8 Allthe faсes of a square-based pyramid of side 6сm.slope at an of 60'to the horizontal. Find the height of the pyramid, and the angle between a sloping edge and the base. A vertical flag pole standing on horizontal ground has six ropes attaсhed to it at a point 6m from the ground. Тhe other ends of the ropes are attaоhed to points on the ground whiсh lie in a regular hexagon with sides 4 m. Find the angle whiоh a rope makes with the ground.

Тhe diagram shows a roof struсture. PQЕS is a horizonta| reсtangle. Тhe faces ABRQ' AB9P' АPQ and BЛS a|| make an angle of 45'with the horizontal.

Find the angle made by the sloping edges with the horizontal.

эql

s.а.

эo

=o eq ogl

ln this сhapter you will lэarn:

.

э ч

. .

howtoextendthe

definit.ons of sine, сosine and tangent to anglos gвatorthan 90o and |ess than 0o the shape of the graphs of sine, оosine and tangent the meaning of the terms .period' and .periodiс'.

5.1 lntroduction

Thеn sin 0.

If, еither by aссident or by ехperimеnting with your сalсulatoц you hаve triеd to find sinеs, сosinеs and tangents of angles outsidе thе range from 0 to 90", you will havе found that your сalсulator givеs you a valuе. But what doеs this valuе mеan, and how is it usеd? This сhapter gives somе answers to those

Thе arrow laЬеlling thе аnglе 0. in Figurе 5.1 еmphasizеs that anglеs arе mеаsurеd positively in thе anti-сloсkwise direсtion, and nеgativеly in thе сloсkwisе dirесtiоn.

questions.

=

} and сos 0o

=

ate thе dеfinitions of sinе and

сosinе foг any sizе of the anglе'с0..

It is usеful tо dividе thе planе into four quаdrаnts, сallеd 1, 2, 3 and 4, as shоwn in Figurе 5.2.

If you think of sinе, сosine and tangеnt only in tеrms of ratios of .oppositе', .adjaсеnt' and .hypotenuse', thеn it is diffiсult to

givе mеanings to trigonomеtriс ratios of angles outsidе thе intеrval 0 to 180" _ aftеr all, angles of trianglеs havе to liе within this intеrval. Howеver, the definition of tangеnt givеn in Sесtion 2.3 and the dеfinition of sinе and сosinе in Sесtion 3.2 both extend naturally to anglеs of any magnitude. It is сonvеniеnt to start with the sinе and сosinе.

flgutо52

5.2 Sine and cosine for any angle

Thеn for аny givеn anglеs suсh аs 60o,210" аnd -140", you сan assoсiatе a quadrant, namеl5 thе quadrаnt in whiсh thе radius сorrеsponding to the angle liеs.

In Seсtion 3.2,the following сonstruсtion was given as the basis of thе dеfinition of sine and сosine.

Draтy a сirсle with radius 1 unit, and сentrе at thе origin o. Draw thе radius oP at an angle 0" to thе x-axis in an аnti-сloсkwisе dirесtion. See Fфrе 5.1. Let P havе сoordinatеs (r, y). flgшвe3

In Figurе 5.3, you саn seе that 50o is in quadrant 1, and is сallеd a first quadrant anglе; 210" is а third quadrant anglе;

-40" is a fourth quadrant angle. You сan havе аnglеs grеatеr than 360o. For ехаmple, you сan сhесk that 460. is a sесond quadrant angle, and 460" is a thiтd quadrant аnglе.

Thе dеfinition of sin 0", namely sin 0" = y, shows фat if 0o is a first or sесond quadrant anglе, thеn the y-сoordinatе of P is positive so sinoo > 0; if 9o is a фird or fоurth quadrant angle, sinOo < 0. figtш5.1

You сan see from this dеfinition, and from Figurе 5.3, that sin60" > 0, that sin210o < 0 and that sin(-40") < 0; you сan еasily сhесk thеsе from vour сalсulator.

Similarl5 thе dеfinition of сos0o, namеly сosO" = x, shows thаt if 0" is a first or fourth quadrant anglе thеn thе x-сoordinatе of P is pоsitivе so сos 0o > 0; if 0o is а sесond or third quadrant angle, сosOo < 0. You сan also sее that сos60. > 0, that сos210o < 0 and that сos(-40") > 0. Again, you сan еasily сheсk thеse from your сalсulator.

Sine and сos.ne for multiples of 90o

flgurc 5.5

Thе еasiest way to find thе sinе and сosine of angles suсh as

90o, 540" and -90. is to rеturn to the definitions, that is sin0o y and сosOo = x. Sее Figurе 5.4.

=

/Т\ щ:r+, l\/R\ L_@-I, 0 / \0 i (0'1)

у^

(-1,0)

\=-/

You сan sеe that the graph of y = 569o has thе form of a wavе. As it repеats itself evеry 360o, it is said to bе periodiс, with pеriod 360". As you would expeсt from Sесtion 5.2, thе value rlf sin 0o is positive for first and sесond quadrant angles and nеgative for third and fourth quadrant anglеs. F.igurе 5.6 shows thе graph of y = gq5@o drawn for valuеs of

I

from -90 to 360.

\:/

(0,

0

-1)

tigшre 5.4

Thеn you sеe from thе lеft-hand diagram that the radius for 90. ends at (0, 1), so sin90o = 1 and сos90o = 0. Similarly, thе radius for 540o еnds up at (-1' 0), so sin.540o = 0 and сos540o = -1.

Finall5 thе radius for -90" ends up at (-1' 0), so sin(-90") and сos(-90") = 0.

=

-1

onсe again, you сan сhесk thеsе rеsults from your сalсulator.

5.3 Graphs of sine and сosine funсtions As thе sinе and сosine funсtions arе definеd for all anglеs you сan draw their graphs.

Figure 5.5 shows thе graph of у = sinOo drawn for valuеs of 0 from -90 to 360.

figuв 5.6

As you сan sее' thе graph of' у = сosO" also has thе form of a wavе. It is also pегiodiс, with peгiod 360". Thе valuе of сos0" is positive for first and fourth quadrant angles and nеgativе for sесond and third quadrant anglеs. lt is thе wavе form of these graphs and thеir periodiс propеrties whiсh make thе sinе and сosinе so useful in appliсations. This point is takеn furthеr in physiсs and еnginееring.

Еxercise 5.1

ln questions 1 to 8, use your сa|сulator to find the fo|lowing sines and сosines.

1 3 5 7

siп 130.

sin250" sin(-20)'

sin36000'

2 cos 140" 4 сos370" 6 cos 1000'

I

сos(-90)"

ln questions 9 to 14, say in which quadrant the given angle Iies.

I 11

200" (-300)"

13 -6000

10 12 14

370" 730" 1000"

lf thе anglе 0" is a first quadrant anglе, tanOo is positive. For a sесond quadrant anglе, y is positivе and r is nеgativе, so tanOo is nеgativе. For a third quadrant angle, y and х are both nеgativе, so tan 0o is positive. And for a fourth quadrant angle, y is nеgative and x is positivе, so tan0o is negative.

5.5 Graph of the tangent funсtion .lust as you сan draw graphs of thе sinе and сosine funсtions,

you сan draw a graph of thе tangеnt funсtion. Its graph is shown in Figuтe 5.8.

ln questions 15 to 20, find the following sines and сosines without using your сalсulator.

15 cosO 17 cos270" 19 cos(-180)"

16 sin 180" 18 sinфO). 2!l sin450"

5.4 The tangent of any angle In Seсtion 2.3 you saw that thе dеfinition of thе tangеnt fоr an aсutе anglе was givеn by tanO" = }. тьis dеfinition is еxtеndеd to all anglеs, positivе and negativе. Sее Figurе 5.7.

figшrc5.7

llgurв 5.8

You сan sеe from Figurе 5.8 that,like фе sinе and сosinе funсtions, the tangеnt funсtion is periodiс, but rлdth pеriod 180o, rathеr than 360o.

You сan also see that, for odd multiples of 90o, the tangеnt funсtiоn is not dеfinеd. You сannоt talk about tan90". It dоеs

not еxlst.

5.6 Sine, cosine and tangent Thеrе is an important relation bеtwееn thе sine, сosine and tan. gеnt whiсh you сan dеduсe immеdiately from thеir dеfinitions. From thе definitions sinOo = У,

сosO"

=

t^nO

=I,

ff,

you сan seе that tanOo =

Еquation

thе book.

1

will

#

bе used rеpеatеdly

Еquation

1

throuфout thе rеmaindеr of

1 3 5 6 7

Е

пI

Exerсise 5.2 |n

oo дI o

questions 1 to 4, use your сalсulator to find the foIIowing tangents.

2 4

tan120' tan200'

tan(r3O)"

tan1000'

Attempt to find tan9Oo on your сa|сu|ator. You shouId find that it gives some kind of eror message. sin 12o ,rm'.

Calсu|ate the value р1 .

CalсuIate the vаIue .)т

сos 1000o

Бj;тбб6'.

t

П

t! э

o GT a a П!

э

-t з

т' t

o

!n

. .

.

ffis chapter you will learn: how to solve simple equations involving sine, сosine and tangent the moaning of .prinсipaI angle'

how to uso the prinсipal angle to find all solutions of the equation.

6.1 lntroduсtion This сhaptеr is about solving equations of thе typе sin 0o сosOo = 0.2 and tanOo = 0.3.

= 0.4,

lt is

еasy, using a сalсulаtor, to find thе sinе of a givеn angle. It is also еаsy, with a сalсulator, to find one sоlution of аn еquation suсh as sin0o = 0.4. You usе the sin-1 kеy and find 0 = 23.57.... Sо far so good.

Thе problеm is that Figurе 6.1 shows therе arе many anglеs, infinitеly many in faсt, for whiсh sinO" = 0.4. You havе found onе of them - how do you find the othеrs frоm thе anglе that vou hаve found?

Y= sin 0o

у', figшo6.1

Figurе 6.1 shows thаt therе is anothеr anglе lying bеtwееn 90o and 180. sаtisfying thе еquаtion sinOo = 0.4, and thеn infinitеly many othеrs' rеpеating every 360".

flgurв 6.2

So thе quеstion posed in Seсtion 6.1 is, .Given the prinсipal angle for whiсh sin 0o = 0.4, how do you find all thе otЬеr anglеs?'

Look at the sine graph in Figurе 6.1. Notiсe that it is symmеtriсal about thе 90o point on the O-axis. This shows that for any anglе ao sin(90 - d|" =sin(90 + с)". фuation 1 If

youwгitе х =90

-d,

thеn

d =90

-ff'

so 90

d" sino"=sin(180-с)".

Еquation 1 then beсomеs, for any

+

ang|e

а

= 180

-x.

Еquation 2

Еquation 2 is thе kеy to solving equations whiсh involvе sines.

Rеturning to the graph of у = sinOo in Figurе 6.1, and using Еquation 2,уoll сan sее that thе other anglе between 0 and 180 = 0.4 is

with sin0"

6.2 $olving equations involving sines Prinсipa| angles

sin-10.4 = 180 - 23.57... = L56.42... . Now you сan add (or subtraсt) multiples of 360. to find all the other anglеs solving sin0o = 0.4, and oЬtain

Thе angle givеn Ьy yоur сalсulator whеn you press thе sin-l key is сallеd thе prinсipal an8lеs.

0 = 23.57, 1 56.42, 383.57, 576.42,... сorrесt to two dесimal plaсеs.

For thе sinе funсtion the prinсipal аngle liеs in thе intеrval

180

-

-90<0<90.

Summary

If you draw thе graph of y = 5i1-t x using your сalсulator valuеs you gеt thе graph shou,n in Figurе 6.2.

To solvе an еquation of thе foгm sinffo = с rмherе с is given: о find thе prinсipal angle

о о

use Еquation 2 to find anothеr anglе for whiсh sinffo = с

add or subtraсt any mulфle of 360.

Example 6.1 Solvе the еquation sinffo val from -1.80 to 180.

=

-0.2, giving all solutions in thе intеr-

The prinсipal anglе is -11.54.

From Еquation2,180

-

bеtwееn -180 and 180 by subtraсting 360.

Thеrеforе thе required solution is Thеrеforе thе solutions arе -168.46 and -11.54.

l

Еxample 6.2 360.

2х. Thеn you havе first to solvе for y thе еquation sin y" = 0.5. Notе also that if .r liеs bеtwееn 0 and 360, thеn y, whiсh is 2x, liеs bеtwееn 0 and720. =

Fтom Еquation 2, 180

- 30

Adding multiples o{ 360 tions for y.

Thus so

=

0.5 is sin-10.5 = 30.

shоrлrs thаt 390 and 510 arе also

solu.

150,390' 510

Find thе smallеst positive root of thе еquation sin(2r

+

50).

=

0.1.

Substitutе У = 2х + 50, so you solvе sinyo = Q.t.

As x

)

=

2х

+

50 yоu find that *

0 for a positivе solution,

Thе prinсipal anglе solving sinyo

From Еquation 2, 180 grеatеr than 50.

-

= }{y

*0-

- s0).

so)

)

0, so y > 50.

0.1 is 5.74.

5.74 = 774.26 is thе first solution

Asr=r1(y-so), х =\1tт+.zв - 50)

=

sinOo sinOo sin 0o sinOo

2 4 6 8

0.3 = 0 = = =

-1

-0.45

62.1'3..

sin0"

=

sinO" =

0.45 'l

sinOo = -0.1 sinOo = -0.5

In questions 9 to 16, find the solutions of the given equation in the intervalfrom -180 to 180.

10 sin0' = -O.5 12 sinOo = 1 14 sin0o = 0.9 16 sinO' = -0.766

sinO'= -0.15 sin0"

sinO"

= =

0

-1

sinOo = -0.9

In questions 17 to 24, find the solutions of the given equation in the intervalfrom 0 to 360.

17 sin2xo = 0.5 19 sin3Oo = 0 21 sinl0'= 0.5 23 sin30" = -0.5 25 Тhe height h in metres

26

18 sin20'= 20 sin20'=

0.45

-1

2I2 sin|Ф =t }sin2s<" =2

24

of the water in a harbour t houв after the water is at its mean level is given by h = 6 + 4 sin(3Ot)'. Find the first positivo value of t for Whiоh the height of the water first reaches 9 metres. Тhe length / hours of a day in hours t days after the beginning of the year is given approximately by

h=12-о

15,75,'J,95,255.1

Еxаmple 6.3

From y

7

= 150 is also a solution.

У =2х = 30,

r

=

5

11 13 15

Solvе thе еquаtion sin2xo = 0.5, giving аll solutions from 0 to

Thе prinсipal anglе for thе solution of siny.

з

9

797.54-360=-168.46.

!

ln questions 1 to 8, find the so|utions of the given equation iп the intervalfrom 0 to 360.

I

(-11.54) = 1'97.54 is also a solution, but this is outsidе thе rеquirеd range. Howеvеr, as you сan add and subtraсt аny multiplе of 360, you сan find thе solutiоn

Stаrt by lеtting

Еxerсise 6.1

"o"

(ffit)..

Find the approximate number of days per yeаr that the length of day is longer than 15 hours.

6.3 Solving equations involving Gosines To solvе an equation of thе form сosOo = 0.2 it is helpful to look at thе graph of the соsinе funсtion in Figurе 6.3.

Summary To solvе an еquation оf thе form сosOo

о o о

= с

whеrе с is givеn;

find thе prinсipal anglе use Еquation 3 to find anothст anglе for whiсh сosOo = с

add or subtraсt any multiplе of 360.

Exаmple 6.4

figшв6.3

For thе сosinе the prinсipal anglе is in thе intеrval 0 < 0 < 180. For thе еquation сos0" = 0.2, thе prinсipal anglе is 78.46... .

If yоu draw thе graph of у = сos-1 x using your сalсulator

Solvе thе еquation сos0o = -0.1 giving all solutions in the intеrval -180 to 180, сorrесt to two dесimal plaсеs.

For сos0o - -0.1, the prinсipal anglе is 95.74...

.

Using Еquation 3, thе othеr angle in thе rеquirеd interval is -95.74... . The solutions' сorrесt to two dесimal plасеs, are -95.74 and 95.74.4

Еxample 6.5 Find all thе solutions of the еquation 2 сos 3х" = 1 in thе intеr-

val 0 to 360.

The equаtion 2 сos3*" = 1 саn bе written in the form сosyo = }, whеrе y = }х.Ifx liеs in thе intеrvаl0 to 360, thеn y, whiсh Ь 3x, liеs in thе intеrval 0 to 1080. Thе prinсipal anglе fоr the solution of сos3l" - } is сos_l Ь= 60. From Еquаtion 3, -60 is also a solution of соsyo = ].

figшв6.4

valuеs, you gеt thе graph shoтvn in Figuге 6.4.

Thе symmetry of thе сosine gгaph in Figure 6.3 shows that

сos(-9.)

= сos@o.

Еquation 3

If you usе Еquation 3, you find that сos(-78.46...)o = 0.2.

If thе interval for whiсh vou nееd thе solution is from -180 to 180, you have two solutions, -78.46... and78.46... . [f you need solutions bеtween 0 and 360, you сan add 360 to the first solution and obtain 78.46 aлd281.54, сorreсt to two dесimal plaсеs.

Adding multiplеs of 360 shows that 300; 420, 660,780 and 1'020 arе also solutions for y.

Thus

so

У

=3х

=

60,300, 42o,660,78o,1020

х =20,100' 140' 220,260,340.

l

6.4 Solving equations involving tangents To solvе an equation of thе form tan 0o = 0.3 it is helpful to loоk аt thе graph оf thе tangеnt funсtion, Figurе 5.8. The symmеtry of thе tangеnt graph shoц,s thаt

tanOo=tan(180+0)'.

Еquation 4

For the tangеnt the prinсipal angle is in thс intеrval -90 < 0 < 90. For the equation tan0o = 0.3, thе prinсipal anglе is L6.70... .

Exerсise 6.2

Using Еquation 4 you find that thе othеr angle bеtwееn 0 and 360 satisfyingtanil = 0.3 is 180 + 1.6.70... = 796.70... .

1 сos0" -3 3 сosс"=i

find the prinсipal angle use Еquation 4 to find anothеr angle for whiсh tan9" = с

the interval-180 to 180 inс|usive.

To solve an еquation of thе form tanOo

=с

where с is givеn:

add or subtraсt any multiple of 360.

Еxamp|e 6.6

Solvе the equation tanOo = -0.6 giving all solutions in thе interval -180 to ].80 сorreсt to two dесimal plaсеs.

For tan0o

-

-0.6, thе prinсipal angle is -30.96...

.

Using Еquation 4, thе othеr anglе in thе rеquirеd intеrval, -180 to 180 is 180 + (-30.96...) =L49.04... .

Thеrеfore thе solutions' сorreсt to two dесimal plaсes, arе -30.96 and149.04.1 Еxаmple 6.7

Find all thе solutiоns of thс еquation tan 10" 0 to 360.

=

_L in thе intеrval

Thе еquation tan }o" = _1 сan bе written in the form tan _L'whеre у = Уo = 10.If 0 liеs in thе intеrval 0 to 360, thеn y, whiсh is j0, Пеs in thе interval 0 to 540. The prinсipal anglе for the solution of tanу"

=

_1 is tan-l(-1)

From Еquation 4, 135 is also a solution of tanyo

=

=

-45.

-[.

Adding multiples of 360 shows that 135, 315 and 495 arе also solutions for v in the intеrval from 0 to 540. Thus so

2 4 6 8

=

5 сos20.=} 7 сos}O. = -0.2 9 сos20o =-o.766

Summary o o o

questions 1 to 10, find all the solutions to the given equation in the interval0 to 360 inclusive. In

y = }0 =

.l'з5'

3|5,495

0 = 90,2L0,330.1

In questions 11 to 16, find

11 13 15 17

-0.3 sin 20o = o.4 tan 9{. = 1 сos2xo

=

|anхo =2 tanp" = -O.5

tan20" =

1

tan Jx" = 1.1

10 tan2xo

=

-0.1

allthe solutions to the given equation

12 tan2xo = -0.5 14 сos fo. = 0.5 16 sin 3f = ю.s

in

Тhe height h in metres of wаter in a harbour above Iow tide is given by the equation h = 14 - 10 cos(3Ot)o where t is measured in hours from midday. A ship сan enter the harbour when the Water is greater than 20 metres. Between what times сan the boat first enter the harbour?

7.1 Notation This сhapter is about finding thе lеngths of sidеs and thе magni. tudеs of anglеs of trianglеs whiсh arе not right-angled. It is usеful to have some notation about sides and anglеs of trianglеs.

,

\,,

\,

figшro 7.1

ln thе triang|e ABC,thе anфs arе сalled А, B аnd Q thе sidеs oppositе thеsе anglеs are givеn thе соrrespоnding lоtrlеr-сase lеtters, а, b and с. In thе triaпglе PgR, thе sides oppositе thе

o oo o a o {r o o s)

angles P' Q and R arе p, q and r.

In this сhaptец thе anglеs will bе measurеd in dеgrеes: thus, in Figurе 7.1, А = 95" арProхimatеly.

1+ Цl J

Thе sidеs may bе mеаsurеd in any units you сhoosе, providеd t}еy аrе аll mеasurеd in thе samе units. In somе of thе diagrams that follow thе units of length arе оmittеd.

7.2 Аrea of

-l

э

-

з

C

o -) ц

jxbasехhеiфt

-t э

э

o.

a triangle

Yоu ате familiar with thе formula foт thе аrea of a trianglе, whethет it is aсutе-anglеd or оbtusеаnglеd.

To find a formula for thе arеa in tеrms of thе sidеs and angtеs of a trianglе, you nееd to сonsidеr aсute- aпd оbtusе.anglеd tri. ln thlз chaptoryou wiIl leam: . how to solve prcblems in triangles whiсh arэ not right.

. .

angled

how to use the sine and cosine formu|ae for a triаngle' and how to flnd the area of a

triangle some applications of the sine and сosine formulae for a triangle.

anglеs separаtеly.

triangle

Case 1: Acute-angled

,

Figurе 7.2 shоws an асute-angled trianglе rпith thе pеrpеndiсular draц,n frоm .Д оnto BC. Lеt the lеnф of this pеrpеndiсulat bе h. Lеt thе area

thе tтiangle Ьe Д, so Ь = |аh. To find h уou сan usе thе lеft-hand riфt.anglеd шianglе.

Thus, so

of'

sinB

=L,

h = csinB.

figure7.2

B

Therеforе

tigurэ

Ь=\аh =lасsinB.

Еquation

1

Alternativеly, you сould find h from thе othеr right-angled triangle,

7.3

Thеrеforе sin(180"

Еquation 2

So, using Еquations 7 and 2 for the arca of' an aсutе-anglеd anglе,

Ь=\аcsinB =\аbsinС.

tri

Еquation 3

Notiсе the symmеtry of thе two еxprеssions in Еquation 3. Еaсh of thе expressions } aс sin B and \ab sinC сonsists of } the produсt of two sidеs multipliеd Ьy thе sinе of the angle Ьetweеn them. It folloцrs from this symmetry that thеrе is a third exprеssion for the area of thе trianglе, namеly lbсsinA.

You сould have dеrivеd this еxprеssion direсtly it in Figure 7.2 you had drawn the perpеndiсular from B to АC, or from C to АB,

Thus for an aсutе.anglеd triangle,

ь=|bсsinA =|саsinB =lаb sinC.

Gase 2: Obtuse-angled triangle Figurе 7.3 shows a triangle, oЬtusе-angled at C, with the pеrpеndiсular drawn from А onto BC. Lеt the lеngth of this

so

=

lаh.

To find h уou can use thе largеr right-angled trianglе. Thus, so

sinB =Ц,

h = сsinB.

-

C) =

l'"

Еquation 4

ь= b sin(180" - C),

Ь=\аh =|absin(780" -

с).

Еquation 5

Еquation 5 looks quitе diffеrеnt from Еquation 4, Ьut if yоu

rесall Еquation 2 from Chaptеr 06, you find

sinC=sin(180'-C). Thеrеfore you сan writе Еquation 5 as

Ь =\аbsin(180" Putting Еquations 4 and

.5

- с) =L,absinС.

togеthеr'

Ь=tасsinB =\аbsinC.

Еquation 6

Notiсе again thе symmetry of thе two expressions in Еquation 6. Еaсh of the exprеssions }aс sinB and L,аb sinC сonsists of } the produсt of two sides multiplied Ьy1he sinе of thе angli bеtrмеen thеm. It follows from this symmetry that thеrе is a third expression for the arеa of the triangle, namely \bc sinА. Thus for an obtusе.angled trianglе,

Ь

pеrpеndiсulat Ьe h.

Lеt thе arca of the trianglе be A, so Ь

=\аcsinB.

Alternativеly, you сould hnd h from thе othеr right-anglеd tri-

o

Ь=|аh =\аbsinС.

t=\аh

anglе,

Ь

sinC=isoh=bsinC,

so

с

=

lbс sin A

=

\са sn B

=

\аb

sin C.

Gase 3: Area of any triangle Thus, the area Ь of any triangle, aсute-angled or obtuse-angled, is

ь=lbсsinA =L,саsnB =tabsinC.

Еquation 7

If you think of this formula as

Arеa

=}

x produсt of sidеs x sine of inсludеd anglе

thеn it is indеpеndеnt of thе lеttеring of thе partiсular trianglе involvеd. The units of area will bе dеpendеnt on thе unit of length that is usеd.

Example 7.1 Find thе аrеa of the triаngle with a = l2crrl, D = 11сm and C

=

53". A

Thе arеa Aсm2 of thе trianglе is

Ь=*аbsinС

=|х12x

ffgшtъ 7.4

11 х sin53.

Using thе sine formula,

sin4Oo

= 52.71.

Thе arеa of the trianglе is S2.7 cmz.J

Example 7.2

Find the area of thе trianglе with b

=

so

=

3.26.

The length of sidе а is 3.26cm, сorrесt tо three siрifiсant figurеs.

A =\bcsgг,А

Somеtimеs you may need to atисk a problem

=|х6х4xsin123. Thе arеa of thе trianglе is 10.1сrлР.

o=*#Т

6сm, с = 4cm and А = |23".

Thе arеа Дсm2 of thе trianglе is

= 10.06.

sin 80o а =Т'

indireсф

Example 7.4

In triangle АBC in Figure 7.S, ang|e C = 100o, b c = 5 сm. Find thе magnitudе of the angle B.

l

l

=4

cm and

7.3 The sine formula for a triangle Thе formula for thе area ol а trianglе,

ь

=

!,b

c sin A = |са sinB = \аb

sn C

lеads to а vеry important rеsult.

figurc7.5

Leavе out the A, and multiply the rеsulting equation Ьу 2, and you find

Using thе sine formula,

bcsnA=саsinB=аbsinC.

If you now dividе by thе produсt аbc,уoaoЬtain

sinА sinB sin C с. т=т=

Еquation 8

This fоrmula is сalled thе sinе foгmula for a trianglе' or morе briefly, the sine formula. Example 7.3

ln triaпglе ABС in Figurе 7.4, ang|e A = 40", anglе B b = 5сm. Find the lеngth of the sidе с.

= 80o and

so

sinB т=Т

sin 100o

sinB

4sin_100o.

=

)

This is an еquation in sinB; thе prinсipal anglе is 51.98". The two anglеs bеtweеn 0 and 180. satisfying this equation are 51.98" and 128.02.. Howеver, as thе anglеs of the trianglе add up to 180", and onе of thе anglеs is 100o, thе othеr angles must bе aсutе.

Thеrеforе

B

=

51.98o.

Anglе B is 51.98" сorr€сt to two dесimal plaсеs.

l

Thе situation that arosе in Еxamplе 7.4 wherc you neеdеd to think сarеfully about thе two anglеs rvhiсh satisfy an еquation of thе form sinOo = ... Oссurs in othеr сases. You сannot always rеsоlvе whiсh solution you neеd.

so

and А2.

one triangle is the aсutе-angled А1BC, whеrе ang|e A, = 53.46". Thе other triangle is thе oЬtusе-angled trianglе А2BC whеre

solutions of thе equation sin 0o = =

4сm. Find

sinА sin40" Т=--Т_

sinА=Y

This is an equation in sinА; thе prinсipal angle is 53.46". Thе two anglеs betwееn 0 and 180. satisfying this еquation arе 53.46" and 126.54'. I

This timе howevеr, thеrе is no reason to rulе out the obtusе сase' so thеre аrе twо possiblе answefs to thе quеstion.

Anglе А is 53.46. ot 126.54. сorrесt to two dесimal plaсеs. Figuте 7.6 shows what is hаppеning.

A'

This situation in whiсh the triangle is not defined uniquely is сallеd thе amЬфous сase. It shows that when you usе the sinе formula to find an anglе it is important to takе aссount of both

7.4 The ambiguous case

Using thе sinе formulа,

plaсes,

Az='1.26.54".

Еxamplе 7.5 shows an ехamplе of this kind.

Example 7.5 In a triangle ABС, ang|e B = 40", а = 5сm and b thе anglе А.

[f you try to draw thе trianglе with thе information givеn, you would draw thе side a of lеngth 5 сm, and draw the 40. at B.Il you thеn draw a сirсle of radius 4 сm with its сеntre at C you find that it сuts the line whiсh you drеw through B in two

..

.

rмhiсh arises in the solution.

Exerсise 7.l

ln questions 1 to 5' find the unknown sides of the triang|e АBC, and ca|сu|ate the area of the triang|e.

1 2 3 4 5

A=54", B = 67. and a = 13.9сm. A= 38.25.' B = 29.63. andb = 16.2сm A =7O", C = 58.27" and b = 6mm A= 88o, B = 36o anda = 9.5оm B =75",C =42" andb =25.0сm

In questions 6 to 9, there may be more than one solution. Find all the solutions possible.

6 7 8 9

b

=

30.4оm' с

=

34.8сm, B =25o.Find

C'А anda.

b=70.25оm'с=85.3сm, B=4o". FindC' Aanda. a =96сm,c= 100cm' C=66o. FindА' Bаndb. a=91 сm,c=78оm, C=29.45o. FindА,Bandb.

7.5 The Gosine formula for a triangle Thе sinе formula is еasy to rеmеmber and easy to usе' Ьut it is no hеlp if you know thе lеngths of two sidеs and the inсludеd angle and цrish to find фe lеngth of thе third side. In Figurе 7.7, suppose that you knour thе lеngths of thе sides а and b, and thе anglе C between thеm, and that you wish to find the length of thе sidе с.

figurt7.6

Thеre arе two сasеs to сonsidеr: whеn anglе C is aсute and when anglе C is obtusе. In both сasеs' thе perpendiсular from А is drawn, mееting BC at D. The length of this perpеndiсular is

h.LettЬe lenф CD

Ьe

х.

-2аbcosC.

с2 = а2 + b2

Тhis formula is сalled thе сosinе formula for a triangle' or morе briefly, the сosine formula.

сDB

Notiсe that thе formula ё = а2 + b2 - 2аb cos C is svmmеtriсal in the letters, d, b, c, and А, B and С. Thеrе arе two other formulaе like it. Thesе threе formulaе are

с+a+

a+

figшrв 7.7

Gase

1:

а2

Aсute.angled triangle

Using Pрhagoras's thеorеm in thе trianglе

Еxample 7.6

lеngth of thе sidе с.

Using thе сosine formula,

Еquating thesе еxprеssions for h? g;ves

ё

b2-х2 =ё-(a-х)2, b2 -х2 = ё _а2 +2ах_х2 с2 =a2+b2-2а.х.

AсD,.r

=

bсos C, so thе еxprеssion foт

ё

= а2 + b2

glvеs

and

l

bесomеs

-2аb cosC.

Gase 2: Obtuse-angled triangle PрhagorasЪ'n.*е;': ACD,

Using

;}.-T**

Using Pрhagoras's theorem in thе triang|e

ABD,

hz=с2-(а+х|2. Еquating thesе еxprеssions for h2 gives

b2-х2= l-(а+х|2, b2-х2= ё-а2-2ах-х2 ё= а2 +b2 +2ах.

that is

or In tтiаnglе АСD,

r

b сos(180. - с). Howеver, from thе graph of y = 665 @o in Sесtion 6.3, сos(180o - C} = - сos C. Thus whеn C is obtusе' эс = -bсosC. Thе expтession for с2 therеfore Ьесomеs =

Еquations9

In a triangle, а = 4cm, b = 7cm and anglе C = 73". Find thе

h2=0-(а-x)2.

In tтianglе

A)

=0+a2-2сасosB | с2 = а2 + b2 -2аbcosCJ

Using Pдhagoras's thеorеm in thе triаng|е ABD,

or

_ 2bccos

b2

AСD,

h2=b2-х2.

that is

= b2 + с2

=

а2 + b2

c2 =42

c

=

+P

-2аbcosC

-2x4x7

сos73"

6.97.

Thе lеngth of the sidе c is 6.97 cm approximately.

l

You сan also usе thе сosine formula to find an unknown anglе if you know thе lеngths of thе threе sides.

Еxample 7.7 Thе thгee sidеs of a trianglе have lengths 5сm, 4сm and 8сm. Find the largest angle of the trianglе. The largеst angle is the angle oppositе the longest side. Using thе сosine formula and сalling the angle 0o, you find 82 = 52 + 42

-2

x.5 x 4 сosOo

64=25 +|6-40сos0o 2з сosOo =-4 0 = 725.10. Thе largest anglе of the triangle is 125.10" approximately.

l

Notiсе that thеrе is no ambiguity in using thе сosinе formula. If thе anglе is obtuse, thе сosine will be nеgative.

Exerсise 7.2

questions 1 to 6' use the сosine formula to find the length of the remaining side.

23

|n

1 2

a = 17.1ClTl, C = 28.8сm' 8 = 108o a = 7,86cm. b = 8.54сm, C = 37.42" с = 17.5сm, b = 60.2сm'4 = 63.67o a = 18.5cm' b = 11.1сП, C= 120"

3 4 б а= 4.31 GП,b=3.87сm'C=29.23o 6 a = 7.59сm; с = 5.67сm,B =72.23" ln questions 7 to 10, find the angles of the triangle.

7 a=2cm,b=3сm,с=4сm 8 a =5.4сm,b =7.1CП, C = 8.3сm 9 a =24сm,b = 19сm, с =26 оm

10 a =2.60сm' b =2.85cm, c = 4.70сm 11 Find the largest angle of th6 triangle with sides 14сm, 8.5сm

12 13

and 9сm. Find the smal|est ang|e of the triang|e with sides 6.4сm, 5.7сm and 8.2сm. Attempt to find the |agest angle of a triangle with sides 5.8сm, 8.3сm and 14.1сm. What happens and why?

ln the rвmaining questions you may have to use either the sine

formula or the сosine formula'

14

Тhe shortest side of a triangle is 3.6 km |ong. Тwo of the angles are 37.25" and 48.4o. Find the length of the longest side. 15 Тhe sides of a triangle are 123m, 79m and 97m. Find its angles. 16 Given b = 5.32cm, с = 6.47сm, A =75.23", find B' C and a. 17 |n a triangle АBC find the ang|e АC8 when c = 9.2cm, a = 5сm

t8

andb=11cm.

The length of the side 8C of a triangle АBC is 14.5 m' angle АBC = 71o, ang|e BAC = 57.. оa]оu|ate thr |engths of the sidеs АC

andAB.

19 ln a quadrilatera| ABCD, AB = 3 m, E}C = 4rП, CD = 7.4m, DA = 4.4m and the angle АBC is 90". Determine the angle АDC. ю Determine how many triang|es exjst with a = 25cm, b = 30сm, аnd А = 50" and find the remaining sides аnd ang|es. 21 The |ength of the longest side of a triang|e is 162 m. Тwo of the angles are37.25" and 48.4". Find the length of the shortest side. In a quadrilateral ABCD, AB = 4.3m, BC = 3.4m and CD = 3.8m. Angle АBC = 95o and angle BCD = 115o. Find the lengЁhs of the diagonals.

2

From a point O on a straight line OX lines Op and Oe of lengths mm and 7 mm are drawn on the same side of OX so that angle XoP = 32" and ang|e ХoQ = 55". Find the |engЁh of PQ. Two hooks P and Q on a horizontaI beam are 30сm apart. From P and Q strings PЕ and QR, 18сm and 16сm Iong respeсtive|y, support a Weight at R. Find the distanсe of R from the beam and the ang|es whiсh PR and QFl make with the beam. |n a triang|e ABc, AB is 5сm |ong, angIe BАC - 55o and ang|e ABC = 48.. Ca|сuIate the lengths of the sides АC and BC and the area of the triangle. 5

24

25 26

27

28

Two ships leave port at the same time. Тhe first steams on a bearing 135. at .|8kmhj and the seоond on a bearing 205" at 15kmh-1. Ca|сu|ate the time that will have elapsed when they are 86km apart. AB is a base line of |ength 3 km, and C, D are points suоh that ang|e BАC = 32.25", angle АBC = 1 19.08o, DBC = 60..| 7., and ang|e BCD = 78.75" . Тhe points А and D are on the sаme side of BC. Find the length of CD. ABCD is a quadrilatera| in whiоh АЕl = 0.38 m, 8C = 0.69 m, AD = 0.42m, angle АBC = 1 09o and ang|e BAD = 123.. Find the area of the quadrilateral.

29 A

weight was hung from a horizontal beаm by two сhains 8 m and 9m long respeсtively, the ends of the сhains being fastened to the same point of the weight, their other ends being fastened

to tha beam at points 10m apart. Determine the ang|es whiсh the сhains make with the beam.

7.6 lntroduction to surueying Thе rеmaindеr of this сhaptеr сontains some examPlеs of praсtiсal uses of thе sinе and сosinе rules in survеying. FormuhЪ for these usеs arе not given, Ьесausе йеrе may Ье i temptation to learn them; thеir importanсе is simply that ihеy use thЪ sinе аnd

соsinе гules.

7.7 Finding the height of a distant objeсt Thrее foгms of this problеm arе сonsiderеd.

1 Тhe point vertiсal|y beneath the top

obiect is accessible

of the

A is thе top of a tall objесt whose hеight Й is аnd B is аt its foot, on thе samе horizontal lеvеl as o. rеquired, As B is aссеssible, you сan mеasuге the horizontal distanсе oB. Call this distanсе d. By using a thеodolite you сan find 0, thе anglе of еlеvatiоn of AB. In Figurе 7.8,

3 By measuring direсtion

a horizonta| distanсe in any

It may not bе еasy to obtain a distanсе PQ as in thе previous examplе, whеrе B' P and Q arе in a straight line. You сan thеn usе the following mеthod. In Figurе 7.10 |et

AB

Ьe the height you nеed to find.

figrtrG78

Thеn so

.^n0. =*

figшrз 7.10

h=dtanФo.

2

The point on the ground veЁiсally beneath the top of the obieсt is not accessible In Figuте 7.9 АB is thе hеight tо bе found and B is not aссеssiblе. To flш'd

AB you сan proсееd

as follows.

From a suitablе point Q, usе a thеodolitе to mеasurе thе anglе 0o.

Тhkе a point P on the samе levеl as B, and mеasurе a horizontal distanсe РQ in any suitablе direсtion. Lеt this distanсe be d.

At P mеasurе angle

ф"

, tЬe anglе of elеvation of А, and angle у

,

At Q measurе angle 0". о In triangle APQ you сan usе thе sinе formula to find АP. o In trianglе APB уou сan usе the sinе formula to сalсulate Й. You сan alsо сalсulatе the distanсеs PB and.QB if you neеd'

them.

7.8 Distanсe of an inaссessible objeсt

PdAB

Supposе you arе at

,<6=iv '\s^

Then mеasure a distanсе PQ, сall it d, so that B, P and Q arе in a straight linе. Тhеn mеasure thе anglе фo. a In trianglе APQ, you сan сalсulatе the lеngth О

Thеn, in tтianglе АQB,уoaсan сalсulatе

P and that at А is an inaссessible objeсt Sфrе 7.11.

rrrhose distanсе from P you nеed to find. Seе

fisf,э7.9

Й.

АQ. flгn7.|1

Mеasurе a distanсе PQ, саll it d, in a сonvеniеnt direсion, and angle 0.. Also measure ang|e ф". In trianglе

APQ, you сan

use thе sinе formula to сalсulatе

АP.

7.9 Distanсe between two inассessible but visible obiects

А and B Ье tп'o distant inaссеssible оbjeсts at thе samе level. Seе Figurе 7.12. Lеt

А

О

с ftшэ7.|3

Мark out and measurе the distanсe РQ (Figurе 7.|3|, ca||ed a Ьase liпe, very aссuratеIy on suitablJground. Thеn selесt a point А and mеasurе anglеs а" and p". You сan thеn сalсulatе thе lеngth AP.

a fgшrв7.12

Меasurе thе lеngth d of a сonvеniеnt basе linе PQ at thе samе lеvel as А and B.

At P mеasure anglеs 0. and ф", and at Q mеаsuгe angles 4o aпd p".

АPQ,уouсan usе the sinе formula to find АQ. Similaф in triangle BPQ you сan find QB. Thеn in trianф АQB, you сan usе the сosinе formula to find AB. In trianglе

Neхt sеIесt anothеr point B and mеasurе the anglеs 0" and ф", and сalсulatе the length АB. You noпr havе enouф information to сalсulatе the arеa of thе

PQBA. BQ and by measuring the angles whiсh BP and BQ makе with |9, yoo сan сalсulatе thе arеa of quadrilateral quadrilatеra|

By joining.

PQBA in a diffеrent way аs a сhесk on your rеsulti. You сan nоw sеleсt a new point C and сontinue by using thе samе methods.

By rеpeating this proсеss with othег points' you сan сreatе a network of triangles to сover a wholе distriсt.

As all the mеasuremеnts of distanсe are сalсulatеd from thе original distanсе d, it is еssеntial that you mеasurе thе basе line

7.1O Triangulation

Thе methods еmployеd in thе last two examplеs arе' in prinсi plе, thosе whiсh are usеd in triangulation. This is thе namе given tо thе method usеd to survey a distriсt, and tо сalсulatе its arеa. In praсtiсе, you nеed сorreсtions to allolлl for sea level, and, оvеr largе areas, thе сurvafurе of thе earth, but ovеr small arеas, the errors arе small.

with minutе aссuraсy. Similarly you nеed to mеasurе the

anglеs еxtrеmely aссuтatеly. You should build in сhесks at еaсh stаge, suсh as adding the angles of a triangle to seе if йeir sum

is 180".

As a furthеr сheсk at thе еnd of thе work ot at аny сonvenient ' stage'-onе of the lines whosе lenф has beеn found by сalсulation, foundеd oп prеvious саlсulations' сan bе used as a basе linе, and the whole survеy workеd baсkwards, finishing with thе сalсulation of йе original measurеd base linе.

Example 7.8 Two points liе due wеst of a stationary ballоon and arе 1000m apart. The anglеs of еlevation of the balloon at the two points are 2L.25" and 18.. Find the hеight of the balloon. А d B

fi$rэ

In Figurе 7.15, thе Ьаllooп is at P, and

figшrc 7.14

[n Figurе 7.L4,|et hеight bе Й mеtrеs.

А

bе thе position of thе balloon, and lеt its

2L.25, Ф = L8 and d = 1000m. Thеn, in trianglе АPQ, thе third anglе is 3.25. 0

7.15

Ф = 33.2, 0 = 21.45, d = 53.42 and d = 1000 m. Using thе sine formulа in triangle ABQ to find АQ, noting thаt ang|е BQA = Ф" - 0" = 33.2" _ 21.45" = L1.75",

=

In triangle

sin

т

AQ

so

18"=_Jbът-' sin 3.25"

[n trianglе

ABQ'

sh21.25"

h=АQsin21.25"=

Ж

siп 1"1..75o

1000 sin

АPQ

tan53.42o

1

1.75' h,

=AQ

so

Ь

=

-

O6 _ 1000sin21.45'

In trianglе

^ 1000 sin 18' r*\J=ltiF. a

so

so

sin21.45"

АPQ, using thе sinе formula,

Q is dirесф bеlow P.

h

fu =|976.

Thе height of thе balloon is approximatеly L976 mettes. I

Exаmp|e 7.9

A balloon is oЬservеd from two stations А and B at thе samе horizontal lеvеl, А bеing 1000m north of B. At a givеn instant the bеaring of thе Ьalloon from А is 033.2" and its angle of еlevation is 53.42", while from B its bearing is 021.45.. Calсulatе thе hеight of thе balloon.

=

AQ

tan53.42o

=

1000 sin 21.45'sin

sin 11.75"

53.42'

= 2419.7.

Thе height of the Ьalloon is therеforе 242a mettes appтoximatеly.l Example 7.10

A survеyor who wishes to find the width of a river mеasurеs along a level stretсh on опе Ьank а linе АB, 150m long. From А the survеyor obsеrvеs that a Post P on thе oppositе Ьank is plaсеd so that angle PАB = 51.33o, and_angle PBА = 69.20". what was the тvidth of thе rivеr? In Figure 7.|6' AB is thе mеasured distanсe, 150m. P is thе post.on thе other sidе of the river. PQ, whiсh is drawn pefpendiсular to AB, is thе width ul of the river. Thе angles а aid B arе 51.33. and 69.20..

A

and B are two points on opposite sides of swampy ground. From a point P outside the swamp it is found that PА is 882 metres and PB is 1008 metres. Тhe ang|e subtended at P bу AB is 55.67". ca|сulate the distanсe between A and B. A arld B are two points 1.8 km apart on a Ieve| pieсe of ground along the bank of a river. P is a post on the opposite bank. lt is found that ang|e PAB = 62" and ang|e PBА = 48o. Ca|сulate the

+t50m+ figurc7.16

To find PQ, first сalсulate АP from triangle АPB.

sin69.20"

Thеn

AP

-

=#

sin (180

-

51.33 150

-

69.20)"

AP= 150sin69.20" ' sin59.4f sin51.33"

Then so |p

=

АP sin51.33o

=

=

fr.

150 sin 69.20" sin

siп59.47"

5

1.33o

= 127.'1,.

Thеrеfore thе width of thе river is |27 mеtres approximatеly. I

Еxerсise 7.3

I

2

A surveyol who measures the angle of elevation of a tree as 32" and then walks 8 m diroсtIy towards the tree, finds that the new ang|e of elevation is 43". Calсulate the height of the tree.

From a point Q on a horizontal plane the angle of elevation of

the top of a distant mountain is22.3". At a point P, 500m turther

away in a direсt horizontaI line, the angIe of elevation of the

3 4

mountain is 16.6'. Find the height of the mountain. Two peop|e, 1.5 km apart' stand on opposite sides of a сhurсh steop|ё and in the same straight |ine with it. From one, the angle of elevation of the top of the tower is 15.5" and, from the other, 28.67'. Calculate the height of the steeple in metres.

A surveyor, who wishes to find the width of a rivel stands on

one bank of the river and measures the angle of elevation of a

high building on the edge of the other bank and direсtly

opposite as 31". After walking 110 m away from the river in the straight line from the building the surveyor finds that the angle of e|evation of the bui|ding is now 20,92". Calсu|ate the width of the river.

width of the river. Тhe ang|e of elevation of the top of a mountain from the bottom of a tower 180 m high is 26.42". From the top of the tower tho angle of e|evation is 25.3.. calсulate the height of the mountain. Two observers 5 km apart measure the bearing of the base of the baI|oon and the angle of elevation of the ba|loon at the sаme instant. One finds that the bearing is 04'lo, and the elevation is 24".The other finds that the bearing is 032', and the elevation is 26.62'. Calculate the height of the balloon. Two landmarks А аnd B are observed from a point P to be in a |ine due east. From a point Q 4.5km in a direоtion 060" from P, А is observed to be due south whi|e B is on a bearing 128". Find the distanсe between А and B. At a point P in a straight road PQ it is observed that two distant objeсtsА and B are in a straight Iine making an angIe of 35" at P with PQ. At a point C 2km along the road from P it is observed

АCP is 50o and ang|e BCQ is 64o. Ca|сulate the distanсe between А and B. 11 An object P is situated 345m above a level p|ane. Two people' А and B, are standing on the pIane, А in a direction south-west of that angle

P and B due south of P. The angles of elevation of P as observed at А and B are 34" and 26. respectiveIy. Find the

distanсe between A and B. 12 P and Q are points on a straight сoast line, Q being 5.3km east of P. A ship starting from P steams 4 km in a direсtion 024.5". Ca|сu|ate: (a) the distanсe the ship is now from the coast-|ine; (b) the shipЪ bearing from Q; (с) the distanсe of the ship from Q. 13 At a point А due south of a chimney staсk, the angle of elevation of the staсk is 55o. From B, due west of А, such that АB = 100 m, the elevation of the staсk is 33.. Find the,height of the staоk and its horizonta| distance from А. Ba АB is a base |ine 0.5 km |ong and B is due west of A. ^t point P has bearing з35.7". The bearing of P from А is 314.25.. Нow far is P from А?

15 A horizontal bridge over a river is 380m |ong. From one end, А, jt is observed that the angle of depression of an objeсt, P, on the surfaсe of the water vertiсaIly beneath the bridge, is 34o. From the other end, B, the angle of depression of the objeоt is 62". What is the height of the bridge above the water? 16 A straight line АB' 115m long, lies in the same horizontal pIane as the foot Q of a сhurоh tower PQ. The angle of elevation of the top of the towor at А is 35.. Angle QAB is 62" and ang|e QBА is 48'. What is the height of the tower'? 1т A and B are two points 1500 motres apart on a road running due west. A soldier at А observes that the bearing of an enemyЪ battery is 295.8o, and at Еl, 301.5o. Тhe range of the gr.ms in the battery is 5km. How far сan the soldier go a|ong the road from А before being within range and what length of the road is within range?

gt

сL TI q)

a э

ln this сhapteryou will.loam:

. .

.

thatyoucanmeаsunean

angle in radians as аn aIternative to degrэes the formulae for length of a сircular arо and the area of a оirсu|ar seоtor

how to convo]t fiom rаdians to degrэes and vice versa.

8.1 lntroduоtion

shows a сirсle with фe samе radius Ьut п,ith an anglе of

\Еho dесided that thеrе should Ье 360 dеgrеes in a full сirсlе, and thereforе 90 dеgrees in a right аngle? It is not thе ansчrer to

1

rаd

at the сеntfе.

this quеstion whiсh is important . it was aсtually thе BаЬylonians - it is the faсt that thе quеstiоn еxists at all.

Someone, somern,hеrе did makе thе dесisiоn that thе unit for angle should Ье thе dеgree as wе now know it. Howеvеr, it сould just have еqually bеen 80 divisions whiсh make a right anglе oг 100 divisions. Sо it sееms worth asking, is there a йst uпit for measuгing anglе? or is thеre a bettei сhoiсе for this

uпit than thе dеgrеe?

It turns out thаt thе аnswer is yes.

A bеttеr unit is thе radian.

8.2 Radiаns

fi$rв8.'l

A radian is thе anglе suЬtendеd at thе сentre of a сirсlе by a сirсular aтс equal in length to thе radius. Sеe Figure 8.1.

Look at thе rеlationship bеrweеn thе lеft. and right-hand diagfams. As thе angle at thе сеntrе of thе сirсlе in thе lеft-hand diagram has beеn mulфliеd by а faсtor 0 to get thе right-hand diagram, sо has the arclеngth. Thе nеw arс lenф is therеforе 0 x the original arс lеngth rсm and therеfore tocm. If you сall thе arс lеngth sсm, then

s=r0. Example 8.1

Еquation

1

o is thе сrntrе of a сirсlе of radius 3 сm. The points A and B liе on its сirсumfеrenсе and anglе AОB = 2 rad. Find thе lenф of thе pеrimеter of thе sеgment Ьoundеd by thе arc АB and the сhord AB. fiшrв&1

Thе anglе of 1 radian is rмrittеn as

1.

rad, but if no units arе

given for anglеs you should assumе that thе unit is radians.

If you arе using radians with а сalсulatоr, you udll nеed to make surе that the сalсulator is in radian modе. If nесessary, look up how to usе radian modе in thе manual.

8-3 Length of a сirсular arс Thе right-hand diagram in Figure 8.2 shoтrs a сirсlе of radius r сm with an angle of 0 rad аt the сеntrе. The left-hand diagram

fisrв8.3

Thе left-hand diagram in Figure 8.3 shoтrs this situation. Thе pеrimеtеr must bе found in two sесtions' thе arс AB and thе

сhord AB.

# "'fu""a

Thе length of thе arс АB is givеn Ьy using Еquation 1,

х2 = 6сm. To find thе lеngth of thе сhoгd АB, drop thе pеrpendiсular from o to AB mеeting AB at \ shown in thе rфt-hand part lеngth

of'

АB

=3

2х

(3

arc

of Figure 8.3.

Thеn

АB

=

2AN

=

sin 1) = J.Q5.

Thе perimetеr is then given Ьy pеrimeter = arс AB

+

АB

= (6 + 5.05) сm = 11.05сm.

This еquation is vеry rarеly used in Praсtiсr. !Иhеn y.ou nееd to сoniert radians to degrеes or viсе vеrsa' usе the faсt that соnvеrsion faсtor. ltr rad = 180. and usе еithеr

l

Thus, for еxamplе, 10" = 10

хfutad= rtп

rad.

8.5 Area of a сirсular seсtor

Figurе 8.4 shows a shaded sесtor of a сirсle u'ith radius r units anglе аt thе сеntrе of 0 rad. Lеt thе arca of thе shadсd "nЪ "n bе А units,. rеgion

8.4 Gonverting from radians to degrees Considеr the сasе whеn thе arc of a сirсlе of radius rсm is aсtually thе сompletе сirсumfеrеnсe of thе сirсle. tn this сasе, the arс lеngth is2ltrrcm. Supposе that thе angle at thе сеntrе of this arс is 0 rad. Then, using Еquation 1, thе lеngth of thе arс is rOсm.

Then it follows that 2пr 2tt rad.

=

r0, so that thе anglе at thе сеntrе is

But as thе anglе at the сеntrе of thе сirсlе is 360",

nd lt nd

2п Thеrеforе

=

360o.

llgшrrc&4

Thе arеa А units2 is a fтaсtion of thе arеa of thе wholе сirсle. As thе arеa оf thе whole сirсlе is ntr* uпitsz, and thе anglе at the сеntrе is 2lt rad,whеn thе angle at thе сеntrе is 0 rad thе shadеd

.e

arеa is a fraсtion

= 18О".

This еquation, It tad - 180", is thе onе you should rеmеmbеr urhеn you nеed to сhange from degrees to radians, and viсе

|п

lз,d= 45o and

!п

rad= 60".

You сan also work out 1 radian in degrееs from thе еquation ltrrad = 180'. You find that 1 rad =

57.296'.

arеa ofthе сirсlе.

Therеforе the aтеа, in units2, of thе сirсular arс is

f,xпР

vеrsa.

[n many сasеs, whеn anglеs suсh as 45o and 60" are givеn in radians thеy are givеn аs multiplеs of z. That is

f;oftЬetotаl

Examp|е 8.2

o

='Рo.

Еquation 2

А and B liе the area ot lзd. Find 2 = АB and thе сhоrd АB.

is tЁе сеntrе of a сiгсlе of radius 3сm. Thе points

on its сirсumferеnсe and anglе AoB thе sеgrnеnt Ьounded Ьy the arс

Fisure 8.3 shows this situation. Thе area of thе sеgmеnt must ЬeЪund by finding thс аrea of the wholе sесtor oАB, and thеn subtraсting thе arеa of thе triangle ОAB.

18 Find in radians the аng|e subtended at the сentre of a сirсIe of radius 2.4сm by a cirоu|ar arc of |ength 11'4сm. 19 Find the |ength of the сirсu|ar arc whiсh subtends an ang|e of 0.31 rad at the сentre of a оirоle of radius 3.6сm.

2

fgurc 8.5

'Thе

20 Find the area of the circular seсtor Whiоh subtends an arо of 2.54 rad at the сentre of a оirсIe of radius 2.3cm. 21 Find in radians the ang|e that a оirоu|ar seсtor of area 20сm2 subtends at thв оentre of a cirсle of radius Scm. A сircular arс is 154сm |ong and the radius of the arс is 252cm. Find the angle subtвnded at the сentre of the оirc|e, in radians and degrees. 2з The angIes of a triangle are in the ratio of 3 : 4 : 5. Еxpress them

arеa Аcm2 of thе sесtor

oАB

is givеn by using Еquatiоn 2,

A =ьРo =tx3? x2 = 9. To find thе area of trianglе OАB, usе the formula tаb sin С, given on page 65.

Thеn area of triangle

oAB

=

|аb sin C =Ьх32 x sin2

=

4.092.

in radians.

24 A сhord of |ength 8 cm divides a оirоle of radius 5 сm into two parts. Find the area of each paft. 26 Тwo сirоIes each of radius 4 сm overlap, and the |ength of their common сhord is a|so 4сm. Find the area of tha оver|apping region.

26 A new five-sided coin is to be made in the shape of figure 8.6.

Thе area of thе shaded sеgmеnt is thеn givеn by arca of'sеgmеnt = area of seсtor = (9

Exercise

- 4.092)

oAB -

сm2 =

arca ot triang|e

oАB

4.908cmz.I

I

questions 1 to 6, write down the number of degrees in eaсh of the ang|es whiсh are given in radians. |n

z*п

ttrп 32o

4Зo 6 4tr

slo

А ffgшrc8.6

ln questions 7 to 12, find the values of the given ratios.

7 9

Jz sinfrz 11 sinQп+f,п) sin

13 Give the p|aсes.

8

сos

}z

10 сoslz 12 sinfп

ang|e 0'2З4 rad in degrees сorreсt to two deсima|

In questions 14 to 17, express the following angles in radians, using fraсtions of z.

14 15' 16 66'

15 17

72' 105"

Тhe point А on the cirсumfersnсe of the coin is the сentre of the arc CD, which has а radius of 2 сm. Simi|arly B is the сentre of the arс DЕ and so on. Find the arоa of one faсe of the сoin.

9..| Introduсtion [n Sесtion 3.2, pages 20 and21,уoll saw that for any ang|e 0oz

sin(90-0)o=сos0o, сos(90-0)u=sin0o, sin20+сos20=1. Thе third of thеsе rеlations is a form оf Pрhagoras's thеorem, and it sometimеs goеs bу that namе. In Seсtion 5.6, page 52, you saw that

tan}

=щ. сos 0

[n Sесtions 6.2 to 6.4, pagеs 55, 58 and 59, you saw that

.т

o o r+ rt + o 1+ aoo o a q)

сг

sin(180-0)",

сos0o

=

сos(- 0).,

In this сhapter' yоu will explorе thеse and other rеlations, as wеll as mее1ing thе nеw ratios sесant' сosесant and сotangеnt.

You will also lеarn to solvе a widеr vaгiеty of trigonоmetriс

ц)

еquations, using thеse rules to help.

Sоme of the rеlations givеn abovе hold whatеvеr units are used

for mеаsuring angles. Еxamplеs аrе sinz 0 + сosz 0 = 1 and smЧ tan0..!Иhеn this is the сasе no units for anglе arе given. сosU =

-l

Howеver, for some of thе relations the anglеs must bе measurеd in dеgrees for thе relation to be truе. This is the сase for sin(90 - 0)o = сos0o and sin(180 - 0)o = sin0". In thesе сasеs,

э

э

dеgrеe signs will Ье used.

1+ Цr J

o

=

tanOo=tan(180+0)".

-

f

{

sin0"

9.2 Seсant, сosecant аnd сotangent ln this chapter you will leam:

. . .

some rclations between the sine and сosine of аn angle the trigonometriс form of PythagoвsЪ theorem the meaning of seсant' cosecant and сotangent.

The three rеlatiоns seсant' сosесant and сotangеnt, usually abbrеviatеd to seс' сosес and сot' are dеfinеd by the rulеs seсO

=

*,

сoseс0

=

*.,

сoto

=

#,

provided сos0' sin0 and tanO arе not zero. ]n t.hе еarly part of thе twеntieth сеntury, tablеs rмerе usеd to find valuеs of thе trigonomеtriс ratios. TЬеre usеd to bе tаblеs for.sе.с, сosес and сot, but thesе have now all but disappеarеd, aдd !f .you want-thеir valuеs frоm a сalсulator, you nеed. to u."

[f yоu substitutе сos20o.= 1 - sin2Oo you obtain an еquation in whiсh еvery tеrm' exсept thе сonstant, is a multiplе of a Power of sin0, that is, a polynomial еquation in sinOo. Yоu сan solvе

this Ьy thе usual mеthods.

3сos20o=1-2sinO" 3(1-sin20o)=1-2sin0o

Thus

the dеfinitions abovе.

You саn writе PрhagorasЪ thеorеm, sin20 ot thesе nеw ratios.

Dividе еvеry tеrm of

+

сos20 = 1, in tеrms

sin20

to

сos20

limilarly, by dividing

sin2

0

+ сos2

0 = 1 by

singo = 1.2L5... or sin9o = -0..5485... . Thе first of thеsе is impossiЬlе. Thе prinсipal angle сorresponding to thе sесond is -33.27. sin2 0 you сan shorм

Example 9.1 =

l. г;nd thе possible values of tanx, sесx and сosесx.

Using Pythagoras's thеorеm, sin2л

f,rs

+

сos2ff = 1, thе valuе of sin

sinr=tffi=t$.

Then, using,"rr,

=

ff,

- {_33.27) = 2|з.27 is also a solution. (See page 55.) But this is outsidе thе rсquired rangе' so subtraсt 360 to gеt

Then 180

2L3.27-36O=-146.73. Thus thе solutions arc -33 .27

а#, ,.."

Notiсе that if you writе sес0o

=

=

aШ

all the solutions iп

thе tеrms in the equa-

сosOo= .r +;ffiТ

.t4

1

сos20o=соsOo+1

сos20o-сosOo-1=0

= } =;. 5

ъft.'сoseсtr

Example 9.2

БЬ

I

tion will involvе сos0o. Thегеforе

.11.

As сoseсx

-1'46 .7 3 .

+ seсOo grving

5 =

and,

Еxаmple 9.3 Solve фе equation сos0o = 1 thе intеrval from 0 to 360.

tanх=#=+=t]. As sесr

0.

sinOo=2!{Й. 6

1'+co*0=Cos€C20. Lеt сosx

=

This is a quadratiс equation in sinO. Using thе quadratiс еquation formula,

so

1'

tan20 + 1 = sес20.

that

- 2 sinOo -2

sin2 0 + сos2 0 = 1 by сos2 0 to obtain

.oЕ-;m=й' whiсh simplifies

3 sinzOo

=

fa

--

сosoo =

t+].r

Solvе thе еquation 3 сos2 0o = 1 _ 2 sin 0o giving solutions in thе intеrval _180 to 1 80.

1j.6 2

= 1.618... or

-0.618.'.

.

Thе first оf thеsе solutions is impossiblе. Thе prinсipal angle сorrеsponding to thе sесond is 128.t7. Then 360

-

128.17

=

231.83 is also a solution. (Sее pagе 59.)

Thus the solutions arc 128.17 аnd 308.17.

I

94

3A 8E o6'

Example 9.4

Solvе thе еquation 2 sec0" = 2 + tan20" giving all thе solutiоns in the interval from -180 to 180.

Exerсise 9

l

Е o

If you usе Pдhagoras's thеorеm in the form tan20. = sес20o - 1 to substitute for tan2 0" alll thе tеrms in thе еquation will

2 3 4

2

Therеforе

5

э o

э o o

involvе sес0".

2sec0" =2+tan20o

o

= 2 + (seс20o

sес20o

-2

sес0" + 1

=

-

6

]')

0

(seсOo-1)2=0 1

сosOo =.ЪiГe =

so

Thus thе solution is 0.

Example 9.5

Solvе the еquation tan9 = -lt and z inсlusivе.

7 9

1

2sin0 giving all solutions

bеtwееn

It is oftеn usеful to writе equations in tеrms of thе sine and сosinе funсtions, bесausе therе are so many morе simplifying еquations whiсh you сan usе.

So

tan9=2sin0 sino"

2 sino. сos t, =

Multiplying both sides of this equation by сosO givеs

sin0-2sin0сos0=0.

Faсtorizing, , so

sinO (1 - 2 сosO)

=

0

sin0=0orсos0=}.

Using thе mеthods of Seсtions 6.2 and 6.3, solvе thеsе two rquations for 0. when sino

=

Thеrеfore 0

0,0

=

= -Itr,

-1I or 0 or itriwhenсoso

-\o, o,tп

ot

п.I

Let tanO. = t, where 0 |ies between 90 and 180. Ga|сu|ate, in terms of t, the values of seс9. сosO" and sinoo. Let seсO = s' Where 0 is aоute. Find the values of cotO and sin0 in terms of s.

сos20"

=

l,0 = -\x

ot \п.

8

= 1

2sin20'-

2sin0'= cosocOo

10 2сos20o = зsinoo + 2 12 sinO. =2сos0o 14 5(1 - сosO") = 4sin20o

sinOo = 0

11 tanO'= Gos0" 13 2seс0" = CoSeCoo 15 4sino" оosOo + 1 =2(sin0.

0=0.

l

of сoseсс.

In questions 7 to 12, solve the given equation for 0, giving your answers in the interval from -1 80 to 180.

sесOo = 1.

Thеrеfoте

Find the value of сosO given that sinO = 0.8192, and that 0 is obtuse. Find the possib|e va|ues of tanO given that сosO = 0.3. Find the possible va|ues of seсO when tanO = 0.4. Тhe ang|e с is acute, and seо с = k. Find in terms of k the va|ue

+

оos0")

10.1 Gompound angles A сompound anglе is an angle of thе form А

+ B or А - B. This сhaptеr is аbout finding the sinе, сosine and tangent of А + B or А - B in terms of the sine, сosinе and tangеnt' as appropriatе' of thе individual anglеs A and B.

Notiсе immеdiatеly that sin(А + B) is not еqual to sinА + sinB. You сan шy this for various angles, Ьut if it wеrе tгue, thеn sin180"

=

sin90o

+

sin 90o

=1+\=2, whiсh is сlеarly falsе. It is diffiсult to give general proofs of formulaе for sin(А + B) and сos(А + B), and this is not attеmptеd in this book. Proofs rмhiсh apply only to anglеs in a rеstriсted rangе arе givеn. Thе

g) э

o o

formulaе obtainеd will thеn bе assumеd to bе truе for all angles.

ц)

t+ fl

o o a aoo C {r

GT П

1O,2 Formulae for sin(A sin(А - B|

т'

сL

B) and

Supposе that anglеs A and B arе both Ьеtwееn 0 and 90.. In Figurе 10.1, thе anglеs А and B arе drawn at thе point Q, and thе linе QN is drawn of lenф Й. PR is pеrpendiсular to QN, and mееts QP and QR at P and R respeсtivеly.

з

э

+

Lеt the lеngths of PQ, QR, PN and tivеly, as shown in the diagram.

NR

be r, p,

х

and у rеspeс-

ln thiз chapter you will Ioarn: how to find the values of

.

B)' cosИ + 8) and B) knowing the values of the sine, cosine and tangent ofA and how to modify thosэ formulaэ for sin(А - B), сos(А - B} and sin(A

tanИ

. .

+

+

I

tanИ.8)

how to find the values of sin2A, cos2А and tan2A

knowing the values of the sine, cosine and tangent of А.

PffiR figum 10.1

The stratеgy for dеriving thr formula for sin(А

+ B) is to sаy that the arеa of triangle PQR is the sum of thе arеas of triangles

PQN

and

RQN.

Sinсe thе formula for arеa of a trianglе is )аb sinC, thе arеa оf triangle PQR is }rp sin(A + B) and of trianglеs PQN and RQN

arc|rh sinА. and |Ph sinв rеspесtivеly.

PQR

Thеn arеa of trianglе anglе

RQN

so

= arca of trianglе

PQN

+ arca of

вi-

\rh sinA +lph sinB. Thеn, multiplying both sidеs of thе еquatiоn by 2, and dividing Ьoth sidеs bу rp gives

lrp

sin(A

+

sin(А

+

B)

=

B)'pr =b^sinА

+ b..

10.3 Formulae for Gos(А cos(A - Bl

+

в) and

Suppose that angles А and B are both bеtwеen 0 and 90o. In Figure 10.3, whiсh is thе samе as Figure 10.1, thе anglеs А and B arе drаwn at the point Q, and thе linе QN is drawn of length Й. PR is pеrpеndiсular to QN, and meеts QP and QR at P and R rеspeсtivеly. Lеt thе lengths of. PQ' 8R, PN and tivеly, as shown in thе diagram.

NR

be r, p,

х

and у rеsPес.

sinB.

Notiсing thatf; =сosB and b = cosA,the formula bесomеs sin(А

+

B)

=

сosB x sinА

+

сosА x sinB.

+

сosА

This equation is usually written as sin(А

+

B)

=

sinА сosB

sinB.

Еquation

Although Еquation 1. has Ьееn proved only for angles whiсh arr aсute' thе геsult is aсtually true for all anglеs

А

А

1

and B

and B,

positivе and negativе. Frоm now on you may assumr this rеsult.

You сan use a similar mеthod basеd on thе diffеrenсe of two arеas to dеrivе a formula for sin(А - B) from Figure 10.2. (You arе askеd to dеrivе this formula in Еxerсise 10.1, quеstion 15.)

figutз 10.3

The stratеgy in this сase is to usе the сosinе formula to dеrive an еxprеssion for сos(А + B). In trianglе

PQR @ + у)2

-

f

* p,

-

2rpcos(A

+

B),

so, simplifying, and using PythagorasЪ thеorem

2rp сos(A+ B)

сos(А

-X..+

You иrоuld then gеt thе formula

- B)

=

sinА сosB - сosA

* p,

-

@+

fl,

After dividing both sides by 2rp уoll obtain

l02

sin(А

f

=* +Pz-х2-2ху-уz _ =1Р х2) * (Р2 - fI -2ху = h2 + h2 -2ху =2h2 -2ху.

RуN figuв

=

sinB.

Еquatioп 2

Notiсing thatb

Ц = CoSАr

+

B\ =Ц -2. rp rp

= cosB,Д

=

sinА

andуo =

sinB, thе

formula bесomеs сos(А

+

B} = сosА сosB

- sinА sinB. Еquation

3

You сan usе a similar mеthod Ьasеd on Figure L0.2 to derivе a formula for сos(А - B). (Yоu are askеd to dеrive this formula in Еxеrсise 10.1, quеstion 16.)

You would thеn get thе formula сos(А

- B)

=

сosА сosB

+

10.4 Formulae for tan(A tаn(A . в)

sinА

+

sinB.

Еquation 4

tan(A

+

вl

=

Sесtion 3.4, page 26, find the eхaсt valuеs of sin 75o and сos 15o.

Using

sin(А

+

ёl

сos4slnЕ сosАсоsB - sinАsinB sinАсosB . соsАsinB

=

sin45o сos30"

+

сos45" sin30.

tan(A +B1 = .tanА+j4ц4. 1-tanАtanB.

- 30.)

сos4.5oсos30o

+

sin45osin30o

-

^'!6

t^'!2.1

Example 10.2

Еquation 5

As with thе formulae for sin(А - B) аnd сos(А - B), you сan usе a similar mеthod to dеrivе a formula for tan(А - B). (You aге asked to derive this formula in Еxеrсise 10.1, question 17.) Yоu would thеn gеt the formula

tаnА- tanB tanАtanB.

- 0).,

с and B Ьe aсutе, and suсh that сosd = 0.6 and 0.8. Calсulаte the еxасt valuеs of sin(с + B) and =

Let thе anglеs

tanА+ tanB =-1-tanAtanB-

1+

сosАsinB

=E"E .*"*

+

сosАсosB sinAsinB dАсo-;в.- сos,ПБв

=

+

A=45"andB=30'.

=

;БасdE--ББsB.

иn(А -ts)

sinАсosB

сos 15o = сos(45o

Nоrv dividе thе numеrator and dеnominator of this fraсtion bv сosА сosB. Thеn

Thеrеfore

=

Тo find сos 15o, you сould еithеr notе that сos0" = sin(90 and thеrеforе сos15o - sin75o, or you сould say that

'i"(l1* (А + B) соs

+u' - sinАсosЦ

B)

=f"E-Ч"tr =*it-

B),

+ сosАsinB - sinАсosB сosАсоsB - sinАsinB.

tan(А

+

sin75.

you find

tanl =дцg. сosU Then

Еxample 10.1 Using thе valuеs of thе sines and сosinеs of 30o and 45o in

and substituting

B) аnd

Yоu сan usе Еquations 1 and 3 tо derive a formula for tan(А starting frоm the formula

10.5 Worked examples

фuation 6

сosp сos(с

+

B).

First you nееd thе valuеs of sinс and sinB. You сan do this by using Pдhagoras's thеorеm in thе form

Thеn As a is aсute,

sin20+сos20=1.

sin20=1-сos2d =1-0.36=0.64. sind is positivе, so sinс = 0.8.

Similarly,

as

B is aсute, sinB

\rт:wB

=

t[1

= =

+

B)

=

+

B)

=

0.64

+

sinАсosB

+

сosАsinB'

+

Д)

0.36

.o'с

=

0.6 x 0.8

=

0.49

+

B)

сosB

-

-0. So sin(с * F) = 1 and сos(с

- sinс sinB

Thеn

-

=

0.|

2 3 4 5 6

+

= 0.5' and ang|es

8) and сos(А + 8).

-

0)o = -sinOo.

А and B are acute,

Use the formuIa for sin(А - B) to show that

- B) when CoSА

=

0.309 and sinB =

0.23, given that angle А is aсute and angIe B is obtuse.

Use the formu|a for tan(А {2 and {З' of tan 75..

+

.

tan64" - tan25o

oт

+

45г =' ]щ. 1 - tanOo'

0.32 where neither А nor B is a first B) and tan(А + B).

+ B} to

- оosА sinB.

сos(А - B)

=

CoSА сosЕl

+

sinА sinв.

17 Use the method of Section 10.4 to prove that

tan(А-B)=ffi.

10.6 Multiple angle formulae From Еquations 1, 3 and 5 you сan deduсе other important formulaе.

In thе formula sin(А + B) = sinA сosB + сos.A sinB, put B

Then

so

--.

А)

=

sinА сosA

sin2А

=

2 sinА

sin(А

+

+

=

A.

сosA sinА

сosА.

Еquation 7

You may somеtimes nееd to usе this formula with 2A replaсed

sin(90-O)o=cosOo.

=

73. sin 12..

f;ъmат;Бz6а. that sin(0 + 45)" = + (sin0o + оos0"). \z

0.

Use the exact va|ues of sine and сosine of 30. and 45. to find

Let sinА = 0.71 and coSB quadrant ang|e. Find sin(А

tanS2o+tan16o - tan 52., tan 1 6"

1

sin(А - 8) = sinА сosB

the exact va|ues of sin 15" аnd сos75".

Ca|оu|ate the value of sin(А

+ sin

16 Use the method of Seсtion 10.3 to prove that

=0ХсosOo+(-1)xsinOo = -sinOo. I

find the va|ues of sin(А

= 1.2 and tan

't5 Use the method of Seсtion 10.2 to prove that

0)o = cos 270o сos 0o + sin 270. sin

Еxerсise 10.1 1 |f cosА = 0.2 and CoSB

.

14 Prove that tan(o

0.49

270 and'B =0. cos(270

.

Flno тne va|ue oт

13 Prove

- 0.8 x 0.6

Еxample 10.3 =

- B) given that tanА

tan(180-0)'=-tanff. sin 52o сos 1 8o - сos 52" sin 1 8o.

Find the va|ue of

.

Use thе formula for сos(А _ B) to show that cos(270

PutА

I

12 F|no тne value

=

B) and tan(А

10 Find the va|ue of сos 73o cos 1 2" 11

-1.

соs(d

+

By using the formula for tаn(А - B)' pюve that

ц[036

0.8 x 0.8 + 0.6 x 0.6

= =

Similaгly

_ g34

0.5.

Then, using thе formula sin(А

sin(с

Find tan(А B =0.4.

find the exaсt va|ue, in terms of

Ьу 0.

Thеn you obtain

sin0= 2sinpсos}0.

Еquation 8

Еquations 7 аnd 8 arе rеally thе samе formula. Usе whiсhevеr form is more сonveniеnt for the proЬlem in hand.

Again, if you put B

A in thе formula

[f you put B =

+

B)

=

tan(А

whiсh simplifiеs to cos

2A

= сos2.A

Еquation 9

- sin2А.

Note thе way of writing сosА x сosА oг (сosА)2 as сos2А. This is usеd for positive Powers' but is not usuаlly usеd for writing powers suсh as (соsА)-l bесause thе notation сos-1r is rеsеrvеd for thе angle whоsе сosinе is x.

You сan Put Еquation 9 into othеr forms using Pythagoras's thеorеm, siл2А + сos2 А = 1. Vriting sin2А = 1 - сos2А in Еquаtion 9 you obtain

cos2A=сos2А-sin2А

- (1 - сos2А) =2сos2A_t cos2A = 2 сosz A - 1. фuation 10 hand, if you put сos2А = 1 - sin2А in Еquation 9 =

so On thе other

you get

сos2А

cos2А=сos2А-sin2А

сos2А

- | -2

Еquation 11

sin2А.

You сan also wгite Еquations 10 and 11 in the forms

and

фuation L2

1-сos2А =2sin2А.

фuation

13

If you write Еquations 9, 10 аnd 11 in half-аngle form, you get

сos0= 2cos2\0and

\

сosO= 1-2sinzL,o.

B)

=

tanA+ tanB 1

- tanАtanB'

tanА+ tanA

1-tanАtanA 2tanA

=т:;П

,2'?n!,.

so

Еqu ation |7

tan2A = L - tanzA

In half-anglе form, this is

^

Eanu=

Еxerсise 10.2

.l

2 3 4

6

Given that sinА

=

ztan\O

Еquation 18

f,t^*g

t, and that А is aоute, find the values of sin

24,cos2A аnd tan2А.

Find sin20, cos20 and tan20 when sinO

=

0.25 and 0 is aсute.

Given the values of sin 45o and сos 45o, use the formulae of the previous seоtions to сa|сulate sin g0" and сos 90.. Given that сosB = 0.66, and that 8 is acute, find the va|ues of

sin28 and сos2B. Given that сosB

0.66, and that B is not aсute, find the vaIues

=

of sin28 and сos2B.

-

1.

I

Given that сos24 = t, find the two possible values of tanА.

I

Prove that sin,10 =

to Given that coso

Еquation 14

tl

Еquation 15

12 Find the value of

Еquation 16

2j,

cos2А and tan2А. Given that sinА - t, and that А is obtuse, find the va|uвs of sin

7 Find the va|ues of 2sin36" сos36" and 2 оos236"

1+cos2A=2cos2A

сosO= cos2\0_si*\0,

+

tan2A

vou oЬtain

5

- sin2А) - sin2А _2 sinzА =| = (1

so

А in the formula

сosА соsB - sinА sinB, сos(А + А) = сosA сosА - sinА sinА, сos(А

you 8et

=

=

Given that сos20

-lryand

сos 2e =

},find sin |0 anасos}e. =

o.2B, find sin0.

ГГ- сoЬ40ъ

/iй.

+]@..

1O.7 ldentities

Exаmple 10.5

It is often еxtrеmely useful to Ье ablе to simplify a trigonometriс еxpression, or to bе ablе to provе that two expressions arе equal for all possiblе values of thе angle or angles involvеd. An еquation whiсh is truе for all possiblе valuеs of the anglе or anglеs is сallеd an idеntity.

Provе thе idеntity.

1+ сosА -'.ЩА r-сosд =ТinА

It is not сlear whiсh side is the morе соmpliсatеd' so use Меthod 2. The аdvantagе with Меthod 2 is that therе is thеn an obvious way to proсеed, thаt is, сhangе the resulting еxpres-

siоn for L}Is

For ехamplе, sin(А - B) = sinА сosB - сosА sinB is an examplе of an idеntity, as are all the formulae given in Еquations 1 to 18. So also is 1 + sin0 = 1sin}0 + сos}o;2, but thе laшеr nееds

- RHS into a singlе fraсtion.

LHs-RHS

=,'i'A, l - сoslt -

To provе that a trigonomеtriс еquation is аn identiry you сan

sin2А-1+соs2А

=0. The last stеp follows from Pщhagoras's thеorеm,

sin2А+сos2А=1.

Example 10.4 Provе the identity 1

Еxample 10.6 Provе thе idеntity

LHS

In the work whiсh follows. LHS will bе usеd to dеnote the lеfthand.Jide of an equation and RHS thе right.hand sidе. s:шrzt,O +

2

snL,O сos}O + cos2L,O

= (sin2}0 + сos2}0) +

2 sin\O сos}0.

Now usе Pythagoras's theorеm, and Еquation 8. Thеn

RHs=1+2sin*Oсos*O

=1+sinO=t"t. Sinсе RHS

- LFIS,

the identity is truе.

l

cosa ф

- sinaф

l

= cos2Ф,

Starting from thе left-hand sidе, and using Method 1,

thе lеft-hаnd sidе.

RHS =

соsА)

ffiГ

Sinсе RHS = LHS, thе identity is truе.

The morе сompliсated sidе is the right.hand sidе, so thе strategy rvill be to usе Mеthod 1 and to prove that this is еqual tо

+

sin2А-(1-сos2А)

Hеrе arе somе еxamplеs. + sin0 = (sin}O + сos}o;2.

slnlt

Тn7.tЕФ

сhoosе onе of two possiblе mеthods.

Меthod 2 |f'уo:u do not seе how to proсeеd with Мethod 1, thеn it may hеlp to take thе гight-hand sidе from thе lеft-hand sidе and to prove that the rеsult is zеro.

1+.сogА

sin2А -( 1 -сosA)(1

to bе provеd to bе an identity.

Мethod 1 Start with thе sidе of the identity you beliеve to bе the morе сompliсated, and manipulatе it, using various formulae inсluding those in Еquations 1 to 18, until you arrivе at thе other side.

.

=

=

i:"!; :#')(сos2 сos2Ф x

- cos2ф Еquation я

=

7

RHS.

гyt-hagoгas,s thеorеm arе usеd in thе sесоnd

",,d stеp of the argumеnt. Sinсе RHS

ф + sin2 ф)

- LFIS,

the identity is true.

l

You must bе сarеful not to usе an illоgiсal argument whеn proving idеntitiеs. Here is an еxamplе of an illogiсal argumеnt.

Example 10.7 An illogical'proof' Provethat2=3.

Then

2=3

If

3 =2. Adding thе lеft-hand sidеs and right-hand sidеs of these еquations gives thеn

=

J.

t

This is oЬviously absurd, so the argumеnt itsеlf must bе invаlid. Nеvеrthеlеss' many students usе just this argumеnt whеn attempting to prove somеthing whiсh is true. Bеware! Howevеr,

if you stiсk to thе argumеnts involvеd with Мethods 1 and 2, you will be safе. Exerоise 10.3 In

questions 't to 7, prove the given identities.

1 sin(А + B) + sin(А - B) = 2 sinА сosB 2 сos(А + 8) + сos(А - B) = 2 сosА сosB с9"Ч,=CoSА-sinА 3 сosА+sInА 4 sin3А = 3sinА - 4sin3А sinА. сosА 2 '- сosА sinА sin2А

1 . coso=melo o ^ ъiБэ.. ъБe "o" Z

,

оosO * сosO

sinф сosф

2sin(0+Ф)

sin2ф

10.8 More trigonometriс equations Sometimеs you сan usе somе of thе formulaе on еarliеr pagеs to hеlp you to solve еquations. Herе are some еxamplеs.

Example 10.8 Solve the equation cos2Oo = sinOo giving all solutions betwееn -180 and 180 inсlusivе.

- 2 sin20" (Еquation 11) and you rшill thеn hаvе an еquation in sinO.

You сan replaсе thе сos20" tеrm by 1

so

Faсtorizing

(2

so

-2

=

sin0o

sin20

=

sinOo

2sinz9o+sinOo-1=0. sinO'- 1)(sin0" + 1) = 0 sinOo = 0.5 or sinOo

=

*1.

Using the mеthods of Seсion 6.2,уo:u сan solve thеse equations

5=5. As this is truе, thе original statеmеnt is true, so 2

t

so

сos20o

for 0.

W,hen sinOo = 0.5, 0 = 30 or 150, and whеn sinO" =

Thеreforе 0 =-9О,30 or 150.

l

-L,0

Exeпвise 10.4

=

-90.

|n questioпs 1 to 10, solve the given вquation for 0, giving your answers in the interval from -180 to 180.

I

sin20o

3 сos20o

5 1-

= =

2

cosOo

4

sin0o сos0o

2sin20o

=

2 sinOo

сosoo

7cos20"=сoS0o 94сos39.=3coso"

2coszх"-1=,1 4 sinOo coS0o =

1

о'*on.*J=!Щ2 = }

8

l0

3sinfl=4sinзOo 2tano.=1-tan2oo

1

1.1 lntroduсtion

If you havе a graphiсs сalсulator avаilable, try drawing

thе graphs of funсtions of the form y = 2sinх + 3 сosx and У = 3 sinx - 4 сos r. Thеse two graphs arе shown in Figure 1 1.1. у 5

y= зsiп

,Y"

4

-

4c0s x"

a

2 1

0

-t

q)

-2

+

-3

5 ao {r o 't

-4 -5

Both graphs havе the сharaсtеristiс wave properties of the sine and сosinе funсtions. They have Ьeen enlargеd in thе y.direсtion, Ьy diffеrеnt amounts' and translated in thе r-diгeсtion, by diffеrеnt amounts.

х

Ф

o o o х

+ зcos xo

figutэ 11.1

-t э

+

У=2sin х.

This suggests that you may be able to writе both of thеsе funсtions in thе form y

з

=

Rsin(r

+

с)o

for suitablе valuеs of thе сonstants R and

R is positivе. ln ilris сhapter you will learn:

. thatthegraphof

. .

y = asinx + bсosx is like the graph of a sine or a cosine how to expross asinx + bcosx in the form B sin(x + с)' and find B and с in terms of a and b

howto usetheform

B sin(x

+

с) in appliсations.

d,, wherе thе valuе

of

This idеa is pursuеd in the nеxt sесtion.

11.2 The form У = asinx

bсosx

+ If you try to фoose thе values of R and a so фat thе funсtion y = R sin(х + с) is idеntiсal with y = аsinх + bcosх, you сan start by expanding sin(r

or

+

a) so that

Rsinr сosс + Rсosr sinс у = (Rсosс)sinx + (Rsina)сosx. У=

If this is thе samе funсtion as у оf r thеn

= сl sin

x

b

+

cos

х for all valuеs

Thеsе еquations' in whiсh сos d and sin d arе both positive, show that с is a first.quadrant anglе, and that

Rсosс=dandRsind=b, сosd=

that is

d=

ftandsina=ft.

Therеforе

You сan intеrprеt thеse two еquations by thinking of а and b as the adjaсent and opposite of a trianglе whiсh has R as its hypotеnuse.

Thеrеforе

R=ч} a@.

Alternativеly, if you squarе thеsе equations and add thе two equations, you gеt

R2(сos2с

that is

R2 = а2 + b2 and R =

+

sin2a)

=

^{

^''l

а2 + b2

Тhеsе thrее еquations,

and

сosс

=

а, * b,.

=

b \'l

Thеrеfore

а2 + b2

R=li + Р а2 +

b2

is usеd to mеan

- 4сosx in

+

с)

.idеntiсally еqual to'.

thе form Rsin(.r +

с) with с in

=

а * 5.36.

sina =

Еquation

Ln'l

а2 + b2

1

Еquаtions 2

enablе you to determinе the valuеs of R and o.

Example 11.1

Ехprеss 2sinro + 3 сosro in the form Rsin(x + d)", whеre thе anglеs arе in dеgreеs.

For the funсtion 2sinro + 3сos = 2 and b = 3. From 'co.4 Еquation 1, R = VТ3. Using this valuе in Еquation 2, сosdo

radians.

.='

3

sinх _ 4сosr

d

whеrе

Дand ^',l

Notе that thе symbol

Example 11.2 Ехpress 3sinx

сosro : r/13 sin(x

d = 56.31..1

whеrе

сosd

Thе valuе of R is always сhosеn to bе positivе.

Thеrеforе R= +[} ap. Then сosd = +and sinс

+ 3

1and sinс = {. )J These equations, in whiсh сosd is positive and sinс is nеgativе, show thаt с is a fourth.quadrant anglе, and that:

а2 + b2, +

sinxo

For the funсtion 3sinx _ 4сos X, 4 = 3 and D = -4. From Еquation 1, R = 5. Using this value in Еquations 2,

R2 сos2d + R2 sin2с = а2 + b2,

so

2

56.31..

=fu^ndsinс" =1fu.

=

= 5 sin(.x +

d)

5.36.1

Example 11.3 Exprеss sinx - 2 сosx in thе form R сos(x

+

d), where R > 0.

This is a differеnt form from thе originаl, but it is not diffiсult to adаpt thе mеthods of the bеginning of this sесtion.

Compaгing thе ехpanded form the foгm sinx - 2сosr, givеs

Rсosr сosd - R sinx sinс with

Rсosd =2andRsinс=1. Squaring and adding, as bеforе, gives

R=Vj andсosl1.1

'=

fr

anо smd

=

Ё.

Thus, as сosd and sind arе Ьoth pоsitivе, 0 is а first-quadrant anglе, and

а

Thus whеrе

=

sinr _ 2 cosх

{5 сos(r

+

Exerсise 11.l ln questions

1 3 5 |n

to 6' write the given funсtion in the form R sin(x + с)..

2 4 6

sin x" + cos xo

2оosxo

+

5sinxo

sin xo - 3сos xo

Ssinx" сos xo

Scos

+

-

xo

12оosf

sin xo - sin xo

questions 7 to 9' give с in radians.

7 8 9

10

Write the funсtion sin x + сos x in the form Rcos(x + с). Write the funсtion sin х + сos x in the form Rcos(x - с). Write the funсtion sin x + сos x in the form Яsinfi - с). By writing the funсtions 7сos x + sin x and Sсos x - Ssin x in the form Еsin(x + с), show that they have the same maximum value.

I1.3 Using the alternative form

Thеrе аrе two main advantagеs in writing somеthing likе sinx + сosr in any one of thе four forms Rsin(r + с), Rsin(x - с), Rсos(x + с) and Rсos(х - с). It enablеs you to solvе еquations easil1 and to find thе

maxi

mum and minimum valuеs of thе funсtion without furthеr

work.

Herе arе somе еxamples.

Example 11.4 Solve thе equation 5 sinx" + bеtwееn 0 and 350.

8

сosxo

=

3 giving all thе solutions

Using thе method of Seсtion 11.2,уou саn write 5sinх"

as

{g9 sin(x

+

57.991".

Thе еquation thеn bесomеs

{89 sin(x + 57.99)" =3

+

you^

57

57.991"

=

#.

rеquirе tЩ .solutions of ,in

.99 and 417 ,99.

""

=

#

Thе prinсipal anglе is 18.54, and thе othеr anglе bеtwееn 0 and 360 is thе sесond quadrant anglе (sее Seсtion 6.2), 180-18.54=161.46.

с)

а - 0.46.| .1

sin(x

Let z = х + 57 .99.Then for valuеs of z bеtwееn

0.46. =

or

+

8сosro

You now have to add 360 to thе first of thеsе to find thе valuе of e in the rеquired intеrval. Thеn the two solutions for z are z = 378.54 and,1,61.46. You find thе solutions for x by substitutin$ Z you find that

or

x

=

х + 57.99, and

- 57.99 = 320.55 х = L61.46 - 57.99 = 103.47. = 378.54

Thus thе sоlutions are 103.47 and 320.55".

l

Example 11.5

Find the maximum and minimum valuеs of sinr - 3 сosr and thе values of x, in radians, for whiсh thеy oссur. !Иriting sinr _ 3 сosr in thе form R sin(r - с) using thе mеthods in Seсtion 11.2 givеs sinr - 3 сosx = {Ш sin(x _ 1.249|. Thе quеstion now bесomеs: Find thе maхimum and minimum valuеs of V10 sin(x - 1'.249) and the values of r for whiсh thеy oссur.

You know that thе maximum of a sine funсtion is 1 аnd that it oссurs when thе ang|e isL,lt. Thus thе maхimum valuе of fiO sin(x - 1'.249| is {Ш, and this oссurs whеn х _ |.249 = )п, that is, whеn х = 2.820.

Similaтly the minimum valuе of a sinе funсtion is -1, and this oссurs whеn thе ang|e is!п. Thus thе minimum value of {10 sin(x - 1,259) is -{Ш, and this oссurs whеn.т _ 1,.259 = }п, thatis, whеn х = 5.96|. I

Еxemp|e 11.6

Solvе the equation 3сos2xo in thе intеrval -180 to 180. Bеgin by writing У

=

-

4sin2хo

=

2 giving all solutions

сosyo

3сosyo _4sinуo

- 4sinyo

Solving thе equation 5сos(y

+

= 5сos(y

+ .53.1.3)".

53.13). - 2 gives

360 satisfying this еquation arе

-293.58, 46.42, 66.42, 293.5 8. 53.13 У

=

-293.58,-66.42,66.42,293.58

=-346.71,_||9.s5,L3.29'240.45.

Finally, dividing by 2 as у

x

=

2x

giv,es

-173.35,-59.78,6.65, !20.22. Notе that фе deсision about whethеr to round up or to round dowп thе final figurе on dividing by 2 was madе by kеeping =

morе signifiсant figures оn the сalсulator.

!

Еxamp|e 11.7 Show that thе еquation 2sinх

4 has no solutions.

+

3сos'r

=

You сan rмritе this equation in thе form

fi3

sin(x

+

al

=

|

for а suitablе valuе of с. You саn thеn rеwritе thе еquаtion in thе fоrm

sin(x+с)=ТТЗ. ^ A.

IL

тiз

> 1, thеrе is no solution to this equation.

+

cos0"

=

2

1

3сosOo-2sinOo=1 -8 сos0o - 7 sinOo = 5

4 6

sinO.

+ r/3

12 sinO"

-

сosOo =

1

5 оosOo = 5

сos20o - sin20.

=

-1

n questions 7 to 12, solve the given equation for 0, giving the value of 0 in the interval-180 to 180 inсIusive. f

7 9

сosO"

+ sinOo =

-1

3сosOo-sinOo=2 cosOo

=-8

8

10

V3 sin0' + cosOo = -1 _2 сos0o - 3 sinOo = 3

12 lзсos2Oo-sin2Oo=-1

In questions 13 to 18, find the maximum and minimum values of the funсtion and the va|ues of x in the interval between -180 and 180 inclusive, for whiоh they oссur.

Using thе methods of Sесtion 6.3, thе anglеs Ьetwееn -360 and

+

sin0"

11 6sin0"-7

The prinсipal angle is 66.42.

y

1 3 5

=)

сos(y+53.13).=0.4.

Thus sо

ln questions 1 to 6, solve the given equation for 0, giving the value of 0 in the interval 0 to 360 inclusive.

2хз the equation beсomes

with solutions for y neеded in thе intеrva| -360 to 360. Vriting 3 сos yo _ 4 sin y" in thе form R sin(y - с)o using thе methods in Seсtion 11.2 gives 3

Exerсise l1.2

l

14 у=3cosxo-4sinxo 13 у=2sinx"-cosxo 15 у =VЗоos2x.-sin2x. 16 у =сos2x"-sin2xo 17 у=3sinxo+4сosxo+2 18 у= ^'t2cos2хo-sin2xo+3

12.1 The first set of faсtor formulae ln Еxетсisе 10.3, question sin(А

B)

+

1, you wеrе askеd to provе thе idеntity

sin(А - B)

+

=

2 sinА сosB.

Thе proof of this idеntity rеliеs on starting rмith thе left-hand sidе and eхpanding thе tеrms using Еquations 1 and 2 of Chaptеr 10 to gеt

and

sin(А

+

B)

sinАсosB

=

+

сosАsinB

sinАсosB - сosАsinB. Adding the left-hand sidеs of these two еquations you obtain sin(А - B)

=

thе rеquired rеsult

sin(А

+

B)

sin(А - B)

+

=

2 sinАсosB,

whiсh will bе used in the rеrмrittеn form

{r

o .I з

2 sinАсosB

rr+ Цr

sin(А

+

B)

+

sin(А

- B). Еquation

1

This is thе first formula of its type. Thеsе formulae enablе you to movе from а produсt of sines аnd сosinеs, to a sum oт diffет-

J

o

enсе of sinеs and сosinеs еqual to it.

Example 12.1 Usе Еquation 1 to simplify 2 sin 30. сos 60o.

{i

2 sin30. сos60o

g)

o r+ oo Е П

=

=

sin(30

+

60).

=

sin90o

+

sin(-30)'

+

sin(30

-

60)"

=1-sin30o =

ц)

1-\

=tr..

In thе middlе of фе ехample, the fасt that sin(4) = any anglе 0 was usеd to сhangе sin(.30)" to sin30o.

.т

!n

. .

this chapter you will loam: how to oxprcss the sum and differэnce of two sines or сosines in an altemative form as a produсt how to do this pюсess in roverse

. howtouseboththo

pкюesses in solving problems.

-

sinO for

If you subtraсt thе еquations

and you

фat

sin(А

B)

=

sin(А - B)

=

obtain sin(А

is

+

+ B) .

2 сosA sinB

sinA сosB

+

сosА sinB

sinA сosB - сosА sinB sin(А - B) = 2 сosА sinB, =

sin(А

+

B)

- sin(А - B). Equation 2

If you had usеd Еquation 2 to solvе Еxample 12.|, уotl would say

2 sin30" сos60o

=

2 сos60o sin30"

=

sin(60

+

30)'-

= sin 90o

-

sin 30o

=1

-

Notе thе following points.

sin(60

o

- 30)'

о In Еquations 3

it is not important whеthеr you А - B оr as B _ A: the еquation сos(_O) = сos 0 for all anglеs 0 ensurеs that сos(B_А)=сos(А_B).

sin3Oo

TWo othеr formulaе сome frоm thе еquivalеnt formulaе for сos(А + B) and сos(А - B),

сosА сosB - sinА sinB сos(А - B) = сosА сosB + sinА sinB.

and

+

B)

and

.W-hen

and

=

2 сosА сosB

-B)

=

- 2 sinА sinB.

сos(А

+

B) - сos(А

+

you rewrite thesе еquations you havе 2 сosА сosB = сos(А + B)

сos(А

+

- B) Еquation

3

- сos(А + B). Еquation 4 Notе thе fоrm of thеsе four equations whiсh are gatherеd 2 sinA sinB

=

сos(А - B)

- B) 2 cosA sinB = sin(А + B) - sin(A - B) 2 cosACoSB = сos(А + B) + сos(А - B) 2 sinA sinB = соs(А - B) - сos(А + B). =

sin(A

+

B)

+

sin(А

Еquation

cos(sum|.

}1sи{s0 +30)

=

}1slniв0)

B)

+

sin(А

+

B), givеs

+

+

sin(50 _30))

sin(20)).

l

=

сos(А

- B) _ сos(А

+

B), givеs

сos20.) _20") -сos(70. +20")) = }(сos(70. = } (сos 50. - сos 90")

sin 70o сos20o = 1

}

12 sin 70"

= } (сos

Еquation 4

2 х сos x sin = sin(sum| - sin(differeпсе|, 2 x сos x сos = cos(sutп| + cos(differeпсe|,

=

Using Еquation 4,2 sinА sinB

Еquation 3

2 x sin Х сos = siл(surп| + sin(dфrenсe|,

-

+

Сhangе sin70o sin20" into a sum.

Еquаtion 2

Thеsе formulae arе often rеmеmbеrеd as

2 x sin х sin = cos(differеnсe|

sin(А

Еxample 12.3

togеthеr for сonvеnienсе

2 sinA сosB

=

sin 50 сos 30 = } 12 sin 50 сos 30)

сos(А -B|

B)

The ordеr of thе right-hand sidе in Еquation 4 is diffеrеnt from thе othеr formulaе.

Using Еquation7,2 sinА сosB

First adding, and then subtraсting, thеsе еquatiоns gives +

о

ЕxаmpIe 12.2 Еxprеss sin 50 сos 30 as thе sum of two trigonоmеtriс ratios.

=

сos(А

and 4

take thе differenсе as

=|-ь=*.

сos(А

.diffеrеnсе, In Еquations 1 and 2, it is importаnt that thе А. is found Ьy suЬtraсting B from

=

50" _ 0)

}сos50..

l

Еxerсise l2.1

In questions 1 to 8, express the given expression as the sum or differenсe of two trigonometriс ratios.

1 3 5 7 9 10

sin30 сos0 cos50" cos3Oo

сos(C + 2D\ cos{2C

+ D)

2 4 6 8

sin35'cos45" оos 50 sin 30

сos60" sin30"

сos(3O + 5D) sin(3C - 5D) 2 sin3А sinА and B = 90-D and simp|ify both |n Еquation 1, putА = 90 sides of the resulting identity. What equation results?

-C

|n Еquation 1, put А = 90 - C and simplify both sides of the resulting identity. What equation results?

12.2 The seсond set of faсtor formulae

Еquations 5 to 8 are oftеn rеmеmЬеred as

Thе sесond set of faсtor formulaе are rеally a rеhash of thе first sеt whiсh еnablе you to vrork thе othеr way round, that is, to writе thе sum or diffеrеnсе of two sinеs or two сosinеs as a produсt of sinеs and сosinеs.

х sin(serпisиrп| х cos(semidifferenсе|, sin - sin = 2 x сos(s emisum| х sin(s erпidiffer eпсe|, сos + сos = 2 х сos(sеmisurп) x cos(semidifferenсe),

Starting from

сos

2sinАсosB

=

sin(А

+

B)

+

sin(А

- B)'

+

B)

+

sin(А _ B)

PutА+B=CandA_B=D.

=

2sinА сosB.

+

sinD

=

D yоu will obtain a

A+B=C A-B=D,

and

С:.D. А=С!D 22 andB= С_D sinC+sinD=2sin С+D z сos l-.

+

sin18"

=

Еquation 5

2sin

= 2

9/,8iu.,

Ч.o,

,J+у.o' Z5fl

sin21.5o сos 3.5".

l

Еxample l2.5

sinC* sinD = 2 sin C; D .o, C 1 D. 2-""2

Еquation 6

Similar mеthods appliеd tо Еquations 3 and 4 givе

z,o'ff.o.f

D'

_ cas70

- сosD =

=

2 sin

30 t^ 70 2 sin -2-2

9*

sin

in70

:-

Lt',

вivеs

3a

sin 50 sin 20. I

Еxample 12.6 Solvе thе еquation sin0" - sin30o, giving solutions in thе intеrval -180 to 180. sin0o _ sin 30o = 2 сos20" sin(_0).

Henсе

Еquation 2 then beсomеs

"inC iD,i,,

sin(sеmidiffereпсe rеuersed|,

sinD - 2 sin

= 2

Еquation 2,Ьу a similar mеthod.

=z

+

х

sin 18o into a produсt.

Using Еquation 5, sinC

сos 30

You сan dеduсе a sесond formula in a similar way from

сosC+сosD=

+

Using Еquation 8, сos С

you сan usе simultanеous еquations to dеduсе that

сosC- сosD

сos =2xsin(sеmisum)

2

Change сos30 - сos70 into a produсt.

Frоm thе equаtions

and

-

=

2sinА соsB.

If you сan rмritе А and B in tеrms of C and formula for thе sum of two sinеs.

Then

sin

sin25"

Thеn thе identity Ьeсomes

sinC

+

Example 12.4 Transform sin25o

writе it the othеr way round as sin(А

sin

- -2 сos20" sinO". сos20.

=

0 or sinOo = 0.

Thе solutions of сos20o = 0 arе _ 135, solutions of sin0" = 0 arе _180, 0, 180.

-

45,45,135 and thе

Thеrеforе thе solutions оf thе original equation arе

Еquation 7

С. Еquation

8

-180,

-

135,

- 45,0,45,135,

180.

1

You сould also havе еxpandеd sin30 in thе form 3sin0 - 4 sin30 and thеn solvеd thе еquation 4 sin30 _ 2 sinO = 0 to gеt the samе rеsult.

ExаmpIe 12.7 Provе that if А

+

sinА sinА LHS

B

+

+ +

C

Еxercise 12.2

ln questions 1 to 6, express the given sum or differenсe as the

= 180. thеn

produсt of two trigonometriс ratios.

sinB - sin C _ tanА tanB sinB + sinC =

1 sin4А + sin2А 3 сos4o-сos20 5 сos47o + сos35o

tslд4tsЩьllщ = (sinА sinB)

+

-

+

sin

C

zsinl(а

+

в)сosl(д

2sin\(A

+

B)сos}(А - B)

zсoslс сos1(д 2сos\C.o'] {а -

_ в) _

zsinlссoslс

+ 2

в) - zsin* B)

+ 2

|n questions 7 to

ссoslс

v^ сos0-сos30 ъin6БйEб.

C

-

+

ln questions 15 to 17, you are given that that each of the following results is true.

sin} C

D

RHS, the idеntity is truе.

l

Notiсе horм in Еxample 72.7,the faсts that сos(90 - 0)o and sin(90 - 0)o = сos0" wеrе used in thе forms сos}{а

+

B)

=

sin}C and sin}(А

+

B) - сos}с.

A

+

B

+

15 sinА + sinB + sinC = 4сos}Аcos}всos}с 16 sinА + sinE} - sinC = 4sin}Аsin1Bсos}C 17 сosА + сosB + сosC - 1 = 4sinlАsln}вsin}с

сos}{а-B)+сos}1а*в)

=

sin0+2sin20+sin30 сos0+2сos20+cos30

12 sin0'+ sin20" = 0 11 сos0o-сos20o=0 13 sinO"+sin20.+sin3Oo=0 14 сosO. +2cos20" +сos30o=0

сos1(д - в) - сosl(д * в)

Lнs

10

questions 1 .t to 14, use the faсtor formu|ae to solve the following equations, giving all solutions in the interval 0 to 360.

2sinАsin =ffi=tanАtanB=RHS. Sinсе thе

sinс+sinf сosс+оosB

In

-.o.l(д-в)-'i',1с - B)

сosА-оosSА sin49o - sin23"

0, use the faоtor formulae to simp|ify the given

sin30'+sin60' сos30.-сos60;

zсoslс(сos1(д - в) - sin}с) 2сos} С(сos trtд _ вl + sin} C) сos} {а

sinSА - sinА

expressions.

sin} Cсos} C

sin] Ссos}

.t

2 4 6

=

sin0o

C = 180". Prove

13..| The cirсumсirсle

The сirсumсirсlе of a trianglе АBC (sее Figurе 13.1) is thе сirсlе whiсh passes through еaсh of the vеrtiсes. Thе сеntrе of thе сirсumсirсle will bе dеnotеd Ьy o and its radius will bе dеnotеd bv R.

o o g) o +o a s) o GT s) + o o 1+

tl

flgure 13.1

To сalсulatе thе radius R of thе сirсumсirсlе, drop thе perpеndi сular from o on to thе sidе BC to mееt BC at N. Sее Figure 13.2.

П

П -l

.т

э

П

f

a.

ln this chapter you will leаrn: . that the сirоumсirсle is the

сirсle which passes through

.

.

the vertices of a triangle that the inсirс]e and the three eсirоles all touсh the three sides of a triangle how to calсulate the radii of

these сirc]es from infomtation about the triangle.

figure 13.2

As

BoC is an isosсеlеs trianglе, bесausе two of its sidеs oN biseсts thе basе. Thеrеfore BN = ]t{C = }a.

the linе

arе R,

In addition, angle BoC, the angle at the сentre of the сirсlе standing on thе arc BC, is twiсе thе anglе BАC' whiсh stands on thе same arс. Thus anglе BoC = 2A, so angle BON = А.

Usingsin2С=1_сos2C, sin2C=

Therеfore, in triangle

so

BoЩ

I

2R

2p = .а_. sln.at You will reсognize that thе right-hand sidе of this formula for R also oссurs in thе sinе formula.

аbс ...-sinА-sinB-sinС.

Еquation

1

Еquation 1, whiсh еnaЬles you to сalсulate the radius R of thе сirсumсirсlе, is what somе pеople undеrstand Ьy thе sinе formula fоr a trianglе, rather than thе shortеr vеrsion on page 66. Notе thаt thе еquation zR

=

4т srnА

would itself givе you a prоof

of thе sinе formula for thе trianglе. For, by symmеtry' thе тadius оf thе сirсumсirсlе must bе a property of the trianglе as a

whole, and not .biasеd' towards onе partiсular vertex. Therеforе, if you had started with another vertех' you would obtain thе formulaе 2R = .b _ or 2R = L.

slnб

sln

-aЬ6' Thus2R=-+=*=-. sln1t slnб sшlL

L

Example 13.1 Find thе radius of thе сirсumсirсlе of thе triangle with sidеs of lеngth 4сm,5сm and 6сm. 5 and с

trianglе, say thе largеst anglе, C, using thе сosinе formula. From thе valuе of сos C, work out thе value of sin C, and thеn usе Еquation 1 to find R. Let' а = 4, b =

= 6. Then саlсulatе an anglе of thе

} = а2 + b2 -2аb cosC' 62=42+52-2х4x5xсosС

Using thе сosinе formula,

whiсh rеduсеs

to

40сos C

sinc= ,'164

Б

of the sinе fоrmula, Еquation Finally, using thе full vеrsion 'g=1.

тсl

SlIlА =a It

Thеrеforе

t-*=#

= 5

сosC=}.

=

-ц.you srnU-'

obtain

1,

.D 6 76 L^=т]т=trт' -в-

Thus thе radius R of thе сirсumсirсle is

fiсm.

l

Еxample 13.2

Thе anglеs A, B and С of, a trianglе arе 50o, 60. and 70o, and thе radius R of its сirсumсirсlе is 10 сm. Calсulatе thе arеa of thе trianglе.

Thе standard formula for the area Ь,of a trianglе is Ь

=

|аb sinC.

You сan hnd а and b from thе full vеrsion of thе sinе formula.

Thus'. from 2p

=

-.аsrnА= L' sln б. а

- 2R sinА

=

20 sin50"

b =2R sinB = 20 sin60'.

and

\аb sinC, Ь=\x20 sin50" x 20 sin60. x sin70o

Using thеsе in thе formu|a tr

=

=

200 sin50o sin60o sin70".

Thе arеa of the trianglе is 200 sin 50" sin 60o sin 70o сm2. Thе full vеrsion of

Ь

г 1 a =;;B = thеsinе formula'й #с

l

r

=

2R,

сan oftеn bе usеd, togеthеr with thе tact tЬat thе sum of thе anglеs of a trianglе is 180o, to provе оthеr formulaе сonсernеd with a triangle. As an examplе, hеrе is a proof of thе сosinе formula using this method. This prооf is сеrtainly not thе rесommendеd proof of the сosinе formula, but it doеs have thе advantagе that yоu do not nееd to сonsidеr thе aсutе-anglеd and oЬtusе-anglсd triangles sеparatеly.

13.2 The inсirсle

Example 13.3

Usе thе sinе formula to pfovе thе сosine formula а2 = b2 + с2

-2bc casА.

Starting vrith thе morе сompliсatеd right-hand sidе, first substitute b = 2RsinB and с = 2RsinC. Thеn manipulatе thе right-hand sidе, kееping symmetry as muсh аs possiblе and using А + B + С= 180 judiсiously, into a form that you сan rесognizе as thе lеft-hand side.

The inсirсlе of a triangle ABC (sеe Figurе 13.3) is thе сirсle rмhiсh touсhеs еaсh of thе sidеs and lies insidе thе сirсlе. Thе сentfe of thе inсirсle will be denoted bv / and its radius will Ье dеnotеd bv r.

+ B + C = 180. by substituting B + C = 180 * А and thеn rесognizing that sin(B + С). = sin(180 - А). = sinА" and that сos(B + C)o = сos(1.80 - А)o = -сosАo.

You usе A

RHS=F+ё-2bccosА" =

4R2sirlBo

=

2Rz(zsirlBo

=

2P

((1'

+

C - 8R2sinB"sin CсosАo + 2 жf C - 4sinB"sinCсosАo)

4R2sirf

сos 2Bo) + ( 1

-

-

сos

2C ) - 2

cos A" (2

flguro 13.з

sinB" sin C))

To сalсulatе thе radius r of the inсirсlе, lеt thе pеrpеndiсulars from I on to the sidеs of thе trianglе АBC meet thе sidеs at L, М and N. Join IА,IB and IС. Sее Figurе 13.4.

=2p(2- (сos28. + сos2C) _2cosА"(2 sinB" sinC)) = 2Rz (2 - (2 сos(B + C).сos(B - C).l 2 сosАo(2 sinB" sin C)) =2P(2

+

2сosАoсos(B _ C|" _2сosА.(сos(B -C)" - сos(B

=2l€(2

+

2соsАoсos(B _ с)" -2сosАo(сos(B _с)"

=2P(2

+

2сosАoсos(B - C)o - 2сosАo сos(B _C). - 2сo*А.)

+

+

C).))

А

сosАo))

=2R2(2sir*A"l = (2RsinA')2 =

d =lllS.

Sinсе RHS

= LHs, thе idеntity is truе.

DV

l

f,gure 13.4

Еxerсise l3.1

1 2 3 4 5

Ca|culate thr radius of the сirсumсirc|e of the triang|e А8C given = 10cm and thatangleА = 30.. Find the exaсt radius of the сirсumcirс]e of а triangle with sides 2 сm, 3сm and 4сm, |eaving square roots in your answer. Тhe area of a triangle А8C is 40сm2 and ang|es B and C are 50" and 70o respeсtively. Find the radius of the сirсumсirсle. Prove that а = bоosO + ссos8.

thata

Lets

= }1a +

b

+

с). Provethat s = 4Rсos}Асos.l

L

вcos}с.

Using the formula } ba'е х hеight, thе arеa of the trianglе B/C is

zаr. Similarly, the areа of thе trianglеs CIА and

AIB are

L,br andL,сr.

Adding thеsе you gеt A' thе arca of thе triangle АBC. Therеforе

Ь=tаr *tb, *\"' .-а+b+с

=rх_-.

Dеnoting the exprеssi on you find that

o*

Lч*

,

Ьy s, (fоr sеmi-pеrimеtеr),

/S=4.

Thе сentrе of the есirсlе oppositе А will bе denotеd Ьy Io and its radius will bе dеnotеd Ьy rд.

Еquation 2

Еquation 2 is used to dеrivе rеsults сonсerning thе inсirсlе of a triangle.

Example 13.4

The sidеs of a triangle ate 4 сm,5 сm and 6 сm. Calсulatе thе radius of the inсirсlе. Let. а = 4, b =

5 and c

= 6. Usе thе сosinе formula tо сalсulate

the сosinе of thе largеst anglе, C. Thеn find sin C, and usе this in thе formu|a \аb sin С to find thе area of thе trianglе. Thеn usе Еquation 2.

Using thе сosinе formula, с2

= а2 + b2

_ 2аb сos C, givеs

36=16+25_2x4x5сosC whiсh lеads

to

.o'C

=

*.

Using sin2 C = 1 _сos2 C shows that

E _т sinC-=J,-

figure 13.5

rcт -3\т ]64 =-Г.

64= Using thе formulа \аb sinС for arca,you oЬtаin

A=}х4x5x .E+

For this

trianglе

'=

Ч-=ЦF.

+

Thеrеforе, using Еquation 2, rs

To сalсulate rд, lеt the points of сontaсt of thе есirсlе with the sides of trianglе АBСЬe P, Q and R.

=o

*

}.*

u

=E.

Thе radius

=Ч. of thе inсirсlе is }t'7сm. l

13.3 Тhe ecirсIes

PB

= ,с

and

с+х=b+у.

= Ь,,

х1'} 1'5lт "'24 -

so

PС = У. Thеn RB = x and QC = y. Thеn, from Figure 1.З.5, x + У = а and as tangеnts from an еxtеrnal point, in this саsе А, arе еqual Let.

Solving thе еquations

Xtу=4

х_у=b_с

,

Thе есirсlеs of a trianglе АBC are thе сirсlеs whiсh touсh еaсh of the sidеs and lie outsidе thе сirсlе. There arе thrеe suсh сirсlеs. Figurе ]'3.5 shows part of the сirсle oppоsitе thе vertеx А.

simultanеously' givеs *

=

а+b-С

Ч=3

Notiсе also that АR

=

=

s

-

's.. = а_b+ сandy

с

+

х

=

tr'

с+ (s_ с)

=

s

=

s-

b.

andthatАQ

= 5.

Lоoking аt arеas' and notiсing that thе area A

ARIAQ _ atea BR/аP

= xг€Д

- 2x

arca АR/А _

2xlrдs -2хlro$ =rа(s_(s*с)-(s-b)) =rд(b+c-s)

Thеreforе

-

of trianglе

АBC is

=

area СQIAP

c)

_2x\ro$ -

2s(2(s * а|)(2(s

b|

Thеrеforе

t

=

A numЬеr of thе formulae in

thе previous sесtion involvеd thе sеmi-pеrimetеr of a trianglе, that is, s. Thеre is a vеry famous formula, сallеd Hеron's formula, involving s for thе area af a triangle. It is A ={s(s _а)(s_b)(s_ с),

Thе mеthod of estaЬlishing Heron's formula сomеs from firрt using thе сosinе formula to сalсulatе сosА. Thе value of sinА is thеn сalсulatеd using sinzА = 1- сos2-А. Finally this valuе is substitutеd in thе foйula for arеa, ь=\bс sinА. = b2 +

ё _2bс сosA

givеs

-т. Thеn using sin2А = 1* сos2А = (1 + сosА)(1 obtain

- сosА), you

сosА)(1 - сosА)

l, - b2 + с2 _ а2Il. lt

ё_

=

b + c)

b))

а2||zbc _ b, _

ё

{16- r)(rl bXr -

_lр j.

Ц'

_ с)

").

Еqu atton 4

Еquation 4 is сalled Hеron's formula.

You сan use thе four еquations еstablishеd in this сhaptеr, togеthеr rмith the teсhniquе usеd in Ехamplе 73.4,to еstaЬlish most of фе formulaе you nеed.

Example 13.5

Find thе radii of thе thrее inсirсlеs of thе trianglе with sidеs of 4сm,5сm аnd 6сm.

Letа=4,b=5andс=6. Thеn s =t{а

Then

+

b + с|

=t,'

- o =|,s - |

=

}and

i*

I

* a2|

E;TfJ2=t}^rт.

^= Thеrеforе, from A = re! - а|

so

4bzё

-|Ф+d2-а2|(а2_(b_d2| 4b2l

V4,Gl7П' -

^-

A = {ф - aр

Thеrеforе

b2 + с2 _ а2\

2b" Il'- 2b"

|zbc * b, *

-

s

-

c =*.

In Еquation 3, A = rд$ _ а|, usе Hеron's formulа, Еquation 4, to find thе arеa of thе trianglе.

с^"д-bz+с2_а2 -us.l =

=

r ,. . ..

'Ос

13.4 HeronЪ formula: the area of a triangle

= (1 +

-

с|(а

=\bс sinA

Еquation 3

Therе are rеlatеd formulae fot rg and rg.

sin2А

с||(2(s

-

o'r'

rдG _ а|,

Rеarranging thе сosine formula а2

-

+b

4s(s_a)(s _b)(s*с)

а).

Ь=

- а)(а

(b + с + а)(b + с

=7

2xarea BR/А _ 2x area СQIд

=

= rд\s

-

Ь,

Similarly and

=tr'^

,o=$*?=# ,u=$,t=Ч ,"=s,з=Ч.

Thе radii of thе inсirсlеs arе

a,d

#,-,Ч,*

5{

c^.

l

Notiсе that thе arca of this trianglе was found Ьy an alternativе mеthod in Еxample 1'3.2.1.

Exаmple 13.6 Provе that sin *А

=

Usе the сosinе formula to find сo-sА, and thеn usе Еquation 16 in Chaptеr 1.0, сosА = 1' - 2 sina )А, to find sin jА. Rеarranging thе сosine formula а2

сosA

Thеrеforе 7 -2

sir}+o

so

sin2

2

|А

=

b2

= bz + с2

Ц-

_ 2bс сosА givеs

а2.

'

o,

.

Therеforе

2bc_b2-с2+а2 4bс

с||(а _(b

4bс

- b)) _ _(2(s * cl)t2(s 4bс

Thеrеforе

sin }А

=

2 3 4 5 6

8

_u, t.,'*i

.- (а + (b -

1

7

=Ч# =

Exerсise 13.2

-

с)|

For the triang|e with sides of length 2 сm, 3 сm and 4 сm, find the area and the radii of the inсircle and the three eсircles.

Find the radius of the inсirс|e of the triang|e with sides 3 сm' 4 сm with an angle of 60o between them. Prove thatrrnryr"=

А,z'

Prove thatr= 8B sin }А sin lв sin }C. Prove that r" = 4B cos }я сos ,lв sin ]c. ln Examp|e 13.4,2 a formula for sin }А is derived. Use a similar method to derive a formula for cos !,А, and then use both of them to find a formula tor tan\A.

Provetheformulatan

}{в-O

Prove that r = (s - a) tan }А.

=

fficot

}я.

14.1 The equation sin 0

=

sin

s

In еarliеr сhaptеrs you havе always, whеn askеd to solvе an еquation, been givеn an intеrvаl suсh as -180 to 180 оr 0 to 360 in whiсh to find thе solutions. In this сhaptеr thе task will be to find a way of giving all solutiоns of a trigonomеtriс еquation, not just those solutions сonfinеd to a givеn intеrval.

For ехamplе, if you arе givеn thе еquation sin0" = 0.6427... yOu сan immediatеly look up the сorrеsponding prinсipal anglе, in this сasе 40. You сan thеrеforе rеplaсе thе equation sin0o = 0,6427.., Ьy thе equation sinOo = sin 40o and thе prob. lеm оf finding all solutions of sinOo = 0.6427... Ьy finding all solutions of sin 0o

ц) .I

э

fl

э

sin 40o.

In Sесtion 6.2 уou saw thаt if 40 is a solution, thеn 180 - 40 = 140 is also a sоlution, and that you сan add anу wholе numЬеr multi ple of 360 to givе all solutions.

o a GI o o -a g C r+ o r+ q) o o ao {r f,

=

Thus all solutions of sinO"

and

=

sin 40o arе

..., -320, 40, 4oo,

7

60,...

...,-220,140,500,860,...

.

You сan writе thеsе solutions in thе form

э

40

+

360n and

1,40 +

360п.

(z an intеgеr)

Notiсе that thеse two formulaе follоw a pattеrn. Starting from 40 yоu сan writе thеm, in asсеnding оrdеr, as

т

0x180+40

П

2x180+40 3x18O-40

1x180-40

Thе pattеrn works baсkwards also, so all solutions fall into thе

patтеrn

In

.

this сhapter you wil| learn: how to find all the solutions of simp|e trigonometriс equations

о $enerol formulae for these

solutions.

:

-2х 1'80 + 40 -1x180-40 0x180+40 1x180-40 2x180+40 3x180-40

You сan usе this pattеrn to writе аll thе solutions in onе for-

Thе prinсipal angle is

mula as

L80n

+ (-1.\"

еvеn.

You сan geneta|ize this rеsult furthеr. If you solvе for 0 thе еquation sin 0o = sin do to givе the solutions in tеrms of с you would oЬtain 0 = 180n + (-1)"a (n an integer). Еquation 1 + (*1)"a thеrе is no rеason

so

1'40' 360 + \40, ...

sее that thеsе sоlutions arе thе samе sоlutions as thosе in thе middle of pagе 139, but thеy arе in a diffеrеnt order.

Thеrеforе the formula 180n + (-7)"а rмhеrе n is an intеgеr givеs you all solutions of thе equation sin 0o = sin с" providеd yоu havе onе solution a. In radians, thе samе formula givеs + (_1)"а (z аn

as thе solution оf thе equation sin0 =

0 =\пlт + (-1)" 2o _

{-1Y

Ьo

ь7т

(и an intеgег).

I

coS 0

=

0.7660...,tЬat

..., *320,40,400,760,... ..., -400,

40

You сan

- nlt

= nпtr +

f,п

intеgеr)

40,320,680,...

.

Yоu сan write thеsе solutions in thе form

...,-220,-320,\40,40,500,....

0

+

It is сonvеniеnt to сonsidеr thе еquation сos0o = is, thе equation соsOo = сos 40o.

and

цrherе п is an intеgеr givеs thе solutions

-

20

14.2 The equation сos 0 are

0=1.80n+{*1.)1.40

or

Therеfore

\lt.

Using thе mеthod of Sесtion 6.3, thе solutions of this еquation

If thе оriginal angle had bееn 140, thе formula

.,., _З60 + 740, -180 _ 140, 0 + t40,180

So thе solution of the еquation is

У = пIt + (-1)"

40

whеrе (_1), takes thе valuе -1 whеn z is odd and +1 whеn z is

Notiсе that in the formula 1'80п why d should Ье an aсutе anglе.

{z.

+

36Оn

and40

+

360п

(z an integer).

Thе pattеrn for thеse solutions is еasiеr to follow than that for thе sinе funсtion. It is

0 = 360n + 40 (и an integеr). You сan gеnеralizе this further, as in thе сase of thе sinе funсtion. Thе solution оf thе еquation сosOo = сosd" is 0 = 360n + а (n an

Еquation 2

intеgеr).

Еquation 3

In radians, thе samе formula givеs

sinс.

0

Example 14.1 Find ail thе solutions in dеgrееs оf thе еquation si'Oo

= -1. ь = -} -30. (-1),(-30) or altеrUsing Еquation 1, all solutions arc 780n + nativеly, Т80n . (-1. )"30 for integet n. I

+ = 2п1tr

а

Еquatioп 4

(n an intеgеr)

as thе solution of thе еquаtion сosO =

сosс.

The prinсipal solution of the еquation sinOo

Еxample 14.2 Find ail solutions in radians оf thе еquation sin(20 Lеt'у

=

20 +tп, Thеn thе еquation bесomеs siny

=

+

|п)

}Г}

= \^{1.

14.3 The equation tan 0 Considеr thе еquation tanOo

= tan40",

=

tan

q,

Using thе method of Sесtion 6.4, the sоlutions of this еquation arе

..., -320, -1.40, 40, 220, 400, 580,

7

60,...

.

Exаmple l4.4 g""еral solution in radians of the equationtan2х -

An altеrnativе form is

н"J.Ё.

40 + I80n (п anirrteger). Thе solution of thе еquation tanOo 0 = |80п

+

=

Let 2х = y. Thе prinсipal angle for еquation the gеnеral solution for y is

tanflo is

d (n an intеgеr).

Еquation 5

In radians, thе samе formula givеs 7I1т +

x(1п) (и an intеgеr). y = nI !ьп (z an intеgеr).

У = nItr

This is thе samе 2S

q' (z an intеgеr)

Еquation 6

as thе solutiоn of thе еquation tan9 = tand,.

-Гe talу - -l1 }п, so

Thеrеforе,

^'

* =Ьу, х =|nп

*[п

1n

an intеger).

l

ЕxampIe 14.5

Solvе in radians thе equation cos20 = Cos0.

14.4 Summary of results

Using

0 = 1"80п + (-1,|"а, {n an

integet).

Thе genеral sоlution in radians of thе еquation sin 0 0 = ппtr + (-7)"а (n an

iлteger).

Еquation

З60п + d (n an

=

intеger).

Еquation 2

=

2nrt + d (n an integеr).

=

+

d

(n an intеgеr).

Thе general solution in radians of thе equation tan0 0

= nnt + {х {п

an intеgеr).

сos

d is

Еquation 4

Thе gеnеral solution in dеgrees of thе еquation tan 0o 0 = I80n

сos do is

=

Equation 3

Thе gеneral solution in radians of thе еquation соs 0 0

1

= sin d is

Thе gеnеral solution in degrееs of thе еquation сos 0o 0

sinс" is

=

tan d" is

=

Еquation =

=

2n7t

!

0 (n an intеgеr).

Taking thе positive sign givеs

Hеrе is a summary оf thе rеsults of thе prеvious sесtions. Thе gеnеral solution in degrееs of thе еquation sinOo

Еquatiol4, 20

Q = 2n1I (п an intеger),

and thе nеgative sign 30 = 2n7t (z an integer).

=lnп (z sоlution givеn, Q =\nlt (r an

Thereforе thе сomplеtе solution is 0

Е'quation 6

l

intеger), inсludеs, Notiсе that thе Ьy taking obtainеd Ьolution thе whеn z is a multiplе of ihrее, thе positive sign.

Еxample 14.6

Solve in rаdians the еquation sin30 = sin0.

Using Еquatiоn 1, 30

Taking n

- nп * (-7|" 0

eveш, and

5

tand is

an integеr).

30

writing n =

2t'7т

+

=

2rп gives

0, that is, 0

Tаking nodd,andwriting n=2nl 3Q = (2m + 1)tt

(n an intеgеr).

-0,

+

= rnItr.

1givеs

that is, 40 = (2m + 1'lr.

Thus thе сomplеtе solution is

ЕxampIe l4.3

Find thе gеneral solution in degrеes of thе еquation сos0" Thе prinсipal anglе is 150, so thе gеnеIal solution is 0

=

З60п + 750 (n an intеgеr). I

=

1lз.

еithеr 0

= ?nltr

ot 6

=

!12rn + 1 )z whеre rп is an integеr.

l

Example 14.7 Solvе thе еquatiоn сos20o = sin0o giving all solutions in

t'*l

tl

dеgrееs.

- 0). to rеwritе сos20.=сos(90_0)..

Usе thе fact' tЬat sinOo

=

сos(90

GI r

thе еquation as

Using Еquation 3, 20

=

360n

t

(90 _ 0) (п an intеgеr).

Taking thе positivе sign gives 30

=

360n+ 90 or 0

= 120n +

anglе of dеpression Thе аnglе of dеprеssion of B from А, whеrе

30.

B is Ьеlow А, is the anglе 0 that thе linе АB makеs with thе horizontal,. Sее Figurе 15.1.

and thе nеgative sign

0=З60n_90. Thus thе сomplеtе solution is еithеr 0 = 1.20n

+

30 or 0

=

360n- 90 whеrе n is anintеger. I

Comparе this solution with that for Еxamplе 10.8.

Еxerсise 14

ln questions 't to 10 find the general solution in degrees of the given equation.

I

3 5 7 9

"ino"

сos30"

=

-1

tan20

=

tanO"

4 6 8

sin30"

=

sinO"

10

tanO"

= -'1

questions equation. In

2

l{3

=

'1

1

сos}"

-|

sin(20

- 30г

tan(30 + 60).

сos30" sin30"

=

anglе of еlevation The anglе of еlevation of B from А, wherе B is aЬove А, is thе anglе 0 that thе linе АB makеs with thе horizonta|. Sее Figurе 15.2.

-1

=

-]r/З

= GoS2Oo =

cosO"

to 20 give the general solution in radians of the given

11 sin20 = -l 13 tan(20 +ltt)=O 15 сos(20 +f,п]l=g 17 tan30 = оot({) 19 сos30 = sin$п _ 0)

figure 15.1

12 сos30 = 0 14 sin(30 +f,п)=t

figure 15.2

Ьеaring Thе Ьеaring 0 of B from А is the anglе in degrееs Ьеtwееn north and thе linе AB, mеasured сloсkwisе from thе north. Sее Figure 15.3.

п0rth

16 tan()п - 20) = -1 18 сos30 = sin20

Ю

sin20=_sinO

figure 15.3

o сл a в) т

сorrеsponding angles !Иhеn two parallеl lines, АB and CD, arе travеrsеd Ьy а linе XY, thе marked angles in Figurе L5.4 are сallеd сorrеspоnding anglеs, and arе еqual.

figure 15.7

A rhombus is a quadrilatеral whiсh has all four sidеs еqual in lеngth. Sее Figurе 15.8. In а rhombus, thе diagonals rhombus

сut at right anglеs, and Ьisесt еaсh othеr.

figшв 15.4

isosсеlеs triangle An isosсelеs trianglе is a trianglе with two equal sides. In Figurе L5.5, АB

=

АС.

А

/\^ /\ /\ B-с figurо 15.5

whiсh has onе linе of rеflесtivе kitе A kitе is a quadrilatеral .!.5.6, ABCD is a kitе with АC as its axis of sylrmetry. In Figurе

figure 15.8

similar trianglеs Two trianglеs whiсh havе еqual anglеs аrе similar. Thе sidеs of similar trianglеs arе proPortional to eасh othеr. In Figurе 15.9, trianglеs АBC and XYZ arе similar, and

BС СА AB YZ Zх хY -=-=--

, figure 15.9

/

/,\

/х

a \

,/

'\ , ,'/"

\ B\,

slant hеight, of a сonе Thе slant hеight of thе сonе is thе lеngth of thе sloping еdge АB in Figurе 15.10.

figurc 15.6

periodiс, pеriod A pеriodiс funсtion /(x) is a funсtion with thе property ihat thеrе ехists a numbеr с suсh that f(х) = f(х + с) ior Ъll valuеs of r. The smallеst suсh value of с is сallеd thе pеriod of thе funсtion.

Pythagoгas's theorеm Рythagoras's thеorеm states that in a right-anglеd trianglе with sidеs а, b and с, wherе с is the hypotenusе,

с

/хz

symmetry.

cZ = а2 +

b2, See Figure 15.7.

figure 15,10

сos(А

+

B)

=

CosА сosB

11481

- sinА sinB

сos(А

- B)

=

сoSА сosB

+

сл

tan(А

+

tl

C

з з q) ч<

o {r

tan(А

Multiple angles

Relations between the ratios titO=

t"rrO = сos U

i+ ffihagorasЪ

=. GI

o э o з o

+.ф

=. o rfr o rl з Е q) -

o

equation sin20

+

сos20 =

1

|+tan20=sес20

Relations for solving equations sin0"=sin(180-0)' =

сos(-0)

tan9o=tan(180+0)'

Solution of triangles ь=\bс sinА

=

sinА-sinB-sinC

b-

т;;;БrГB

sin2А=2sinАсosА сos2A=сoszА-sin2А cos2A=2coszА-7 cos2А = 1-2 sinz A tanlA - 'L 2tan4 -tantA =

sin(А

+

B)

2 сosA sinB

=

sin(А

+

B) _ sin(А _ B)

2 cosA сosB

=

сos(А

+

B)

2 sinА sinB

=

сos(А - B) - сos(А

sinD

=

z rin C I D.o,

sinc* sinD

=

z

сosD

=

2.o,

сosC_сosD

=

z sin C 1 D,i,, D ] C

сosC

+

+

с

-2bc cosA b2 = с2 + а2 -2са cosB а2 = b2 + с2

с2 = а2 + b2

=

-tanAtanB - tanB

tanA

2 sinА сosB

sinC

\са sinB =\аb sinC

a

- B)

+ =-|tanA tanB

Faсtor formu|ae

L+co*0=CоS€C20

сos0

B)

sinА sinB

- 2ab сosC

Compound angles сosА sinB

sin(А

+

B)

=

sinА сosB

sin(А

- B)

=

sinА сosB - сosА sinB

+

+

+

sin(А

сos(А - B)

2---2

*tff C

- B)

sin

+

B)

C; D

9/

D .o* Cl/'2

D

2--'2

General solution of equations Thе gеnеral solution in dеgreеs of the еquation sin 0o 0 = 780n + (-1)"а (z an intеgеr). The gеnеral solution in radians of thе еquation sinO 0

= пntr +

(_I|"а (и an integer).

= sin do

=

sinс is

1S

Thе general solution in dеgrеes of thе еquation сos 0" 0 = sin 360n +

а

=

2nlт

!

0 = 180п + d (п an intеgеr). Thе gеneral solution in radians of thе еquation tan0 = пnt +

Radians

E tad

Length of arc

d

(п'

an intеgеr).

= 180 dеgrееs

s=r0

Rаdius of eсirсle whеre

-

b) = 16$

-

'=||а+b+с).

HeronЪ formula for areа of triangle A whеrе

s=)@+b+с).

=

r/s(5

tand is Answеrs involving lеngths arе givеn eithеr сorrесt to thrее sig. nifiсant figurеs or to thrее dесimal plaсеs, whiсhеvеr sееms morе apprЬpriatе. Answеrs fоr anglеs arе usually given сorrесt to two dесimal plaсеs. 0.364 5729.578 1.881

18.88" 80.72" 45"

8.36 m 4.69 miles

126m 53.01',36.99'

-l[J-

6115

-

61

с)

1 3 5 7 9

68.20'

2.51 сm 11.59 сm 8.06 оm

63.86'

Еxerсise 3.1 (page 24)

1 3 5 7

I ll

2 4 6 8

0.577 0.852 0.009

63.43'

10 0.01"

12 600 2 038.66"

4 19.54 m 6 21.3 m

I

144m

75.64"

Exerсise 2.3 (page 17)

'=t{а+b+c). = rвb

U'

1 3 5 7 I

/S=Д

- аl

=

э

tando is

Exercise 2.2 (pаge 13}

Radius of inсirсle

rд(s

=

11

Radius of сirсumсircle с b .,p а а,. = = =;i;т Jlffi ;;ъ-

=

g)

сosс is

1 3 5 7 I

д =\Р0

A

=

Еxerсise 2.1 (page 9)

Area of сirсular sector

wherе

ш

а, (n an intеgеr).

Thе gеnеral solution in degrееs of thе еquation tan0o

0

d" is

(п an intеger).

Thе gеnеral solution in radians of thе еquation сos0 0

= сos

66.42'

2.25 cm 3.37 сm 6.85 cm

60.6'l'

0.994,0.111

3.73 сm 48.81" 3.46 сm 24.78" 10 3.60 cm

2 4 6 8

2 4 6 I

10 12

5.47 cm 28.96" 1.73 сm

27.49" 2.68 cm 5.14 сm,3.06 сm

o .т o €

13

i5

17 19

14 38.4з" 16 4.82 cm

36.87', 53.13" 9.32 km 2.87 cm 38.42", 51.58', 7.4O cm

18 7.19 km

Exerсise 3.2 (page 29)

1 3 5

7 9

2 6.66 сm 4 48.81" 6 2.1cm I 67.з8.

73.30'

5'79 сm 9.44 сm 10.03 сm

10 5.38 сm 12 44.20"

19.02'

1.| 35.02.' 2.86 сm

13 55.5 mm,72.7 mm 15 2.66 оm, 1.86 сm, 3.88 17 69.51' .|9 0.681 сm

21

ж

2'60 сm' 2.34 сm 10.1т km, 11'70 km

Exercise 3.3 (page 33)

1

в 5 7

9

10'05 оm 145'42" 10.91 km,053.19" 44.80 сm 25.24 сm

Еxerсise 4 (page 43)

I

3 5

68.20',60.50" 3.68 m

35.26" 54,т4.

7 I 56.31'

Еxerсise 5.1 (page 50)

1

0.766

з -0.940 5 -o.з42 70 93

11

't

15

1

132 170 t9 -1

сm

14 30.500,59.50' 16 44.12",389.8 mm 18 14.7 km,10.3 km 20 |хlзm,tФ,\

2, 3.59'

24 З28'17"' 17.1 km

2 4 6 8

.|0

2 4

4.61cm

4.33 сm 85.06" 8.58 сm 13.20 сm

73.57',78.22" 97.08' 5.20 cm,50.77"

10 35.26'

2 -0.766 4 0.985 6 0.174 80

10 12

1 1

144 160 18

20

1

з

6

-1 1

2 4.577

-1.732 0.364

4 -5.671

7 -0.'176

0.213

Exerсise 6.l (page 57)

1

з

5 7 9

17.46.162.54

2 26.74.'t53.26 490

206.74,333.26

I

0' 180' 360 270

6185.74,354.26

17 15,75,195,255 19 0, 60, 120,'180, 240, 21 60,300 23 90' 210' з30 25 1.62 hours

12 90 14 64.16,115.84 16 -130.00. -50.00 300, 360

Еxercise 6.2 (pаge 61) 1109.47,250.53

3 5 7 I

11 12

1з

210,330

10 -150. -30

*171.37, -8.63

11 -'180,0,180 13 -90 15 -115.84. -64.16

10

612.72",15.76"

8

Еxerсise 5.2(page52|

t8

20

2,

13.37, 72.63,193.37, 256.63 '135,315

180

24 20.91, 69.09, 200.91, 26 About 122daуs 2 вЗ.43'24з.43

4 '153.43,333.43 4',1.41,318.59 6 22.5, 1 1 2.5, 202.5, 292.5 30, 150,210,330 143.18 203.07 70.00, 1 10.00, 250.00, 290.00 87.14, 177.14, 267.1 4, 357.14 -126.27,-s3.73, 53.73, 126.27 _10З.28' _13'28' 76.72, 166.72

I

*168.2.1

'

_101.79' 11.79,78.21

14 -90,90 15 -90,30,150 16 -45 17 4.14 pm and

7.46 pm

Exerсise 7.l (page 69) Тhese answers are given in alphabetiсa| order.

1 2 3 4 5 6

249.09

15.82 cm, 14.73 cm, 94.22 cm2 20.29 оm, 30.36 cm,152.22 cm2 7.18 mm,6.50 mm, 18.32 mm2 5.59 сm, 7.88 cm,22.0О cm2 23.06 cm, 17.З2сm, 192.88сm2

C C

28.93., А = 126'o7",a = 58.15 сm or = 15.t.07", А = 3.93", a = 4'93 оm

=

7 8 9

А

= 51 .31o,

= 145.00o,

B

= 5.55o,

2 4 6

5.30 сm 25.9 сm 7.97 cm 28.96",46.57"104.47" 40.11", 57.90", 81.99' 62.19", 44'44",7З.З7" 10 28.91.,31.99., 119.10. 12 43.84" 13 You get аn error message from the сalсu|ator beсause the сosine of the |argest angle is -1'029' whiсh is impossible. Тhe probIem arises beсause the triang|e is impossible to draw, as the longest side is longer than the sum of the other two sides. 15 52.01',88.05',39.93" 14 5.93

37.77 crn 54.73 сm 2.1'l cm

I

7 9 1t 't06.23"

km

cm

't 7 16 45.17" , 59.60', 7 .25 19 18 16.35 m,13.62 20 Two triangles' 66.82., 63.18" and 1 6.82., 1 1з.1 8. and 9.44o сm 22 21 98.34 m 24 23 3.09 mm 25 3.81 сm, 4.2o cm,7.81 cm2 2в 28 27 4.41km 29 49.46',58.75'

m

Еxerсise 7.3 (page 80)

1 3 5 7 9

15.2 m

276m

889m 3700m 2.88 km

11 500 m 13 72.9 m,51.1 m 15 189m 17 3470m,727om Exerсise 8 (page 88)

160

з 5 7 I

270 135

0.588 0.309

2 4 6 8

56.09' 41.04'

4.5'l hours

8

10

192m

'1.26 km

0.924 0.383

tв

18 2o 22 24 26

{27т

1.12 оm 1.6 anа

2.9O cm2

fп

Exerсise 9 (page 95}

1

4.75

6.72 сm2 0.611 rad,35-01o 23.18 сm2 and 55.36 сm2 3.03 cm2

"t- lPl a+

-=-!' ----l-. ^,!1+t2' ^,!1+t2

+P-

38.17,141.83

13 -153.43,26.57 't5 -60,30,60, 150 0.663,

tз.180 k

1

a -

r/s2-1' s

8

-135' -45' 45, 1з5

2

t6-tt t6-ta т,4

{s2-1

10 -180,0, 180 12 -116.57,63.43 14 -75.52,0, 75.52

Еxercise 10.1 (page 102}

-0.749

Еxpanding gives sin(90 - 0)" = sin90. сos0o - сos90o sin0o = 1 x сos0o -0 x sinO. = CoSOo

3-V3

{3 -

5 0.894, 1.999

1

or2+l3

7 #= 3.oтт,2* =

Using Equаtion 6' tan(180 - 0)o

=

9

ll*rс

+1.077

7 -120, -60, 60, 120 -180,0,30, 150, 180

5.15 km

+n

2

-^,11

8

0.5

-0.5735

4 -0.997 о 3*€o, €Д

546m

4 120 в 72o

25

l 3

0.305 m2

215

13.41

ж |п,tп

11

5.71 m,6.08 m 7.98 оm, 26.32. and 29.93.

12 з.64 km' 315.' 14 1.246Rm 16 63.7 m

17 19 21

I

29.13 сm

2.23 km 1O 2.17 km

12 14

0.966

t5 |п

b = 15"35 сm

Exerсise 7.2|page72|

1 3 5

11 13

А = 88.69., a =109.26 сm or А = 11.31", a = 21.43 cm 128.69"' = А = 61.28", B = 52.72",b = 87.10 cm А = 35.00., 8 = 115.55", b = 143.13 сm or C C

0-tan0'

1ftffi;

=

tan ]

l'sцl

180'-tan0'

+taггiъoтaпЕ;

=-tano"

sin34" 11 tan68"

10 сos61 o 12 tan39"

Еxerсise 10.2 (page l05}

1

'*,*,?

0.553 5 0.992, -0.129 7 sin72",сos72" з

0.484' 0.875'

2

4 6

8

-ц7-ц 25' 25' 1,0

7

-0.992, -0.129 0.5, -0.5

9

Use Еquation 16, сosO = 1 -2 sin2 f,L,to get sin2 Lo =LP and the resu|t follows. Use Equation 1 5, сos 0 = 2 cot2 Ь0 to get сosz Ь0

10 !ь'*ьФ 12 tan2О"

=

-'-щz

6 LHs ,t ,

=-1 +сos0 sin 0

and the result follows.

11

t0.6

2c:s2Ь0

=2 sin}0сosi0= (12 & 8)

Exerсise 10.3 (page 108)

=

ln the solutions to the identities, the reason for eaсh step is given by

plaсing an Еquation number from Chapter 10 in bold-faсed type in braсkets. Where no reason is given, either an aIgebraiо simp|ifiсation or the use of sin2А + сos2А = 1 is involved.

1

2

LHs

sinИ + B) + sinИ _ 8) = sinА сosB + сosА sinB = 2 sinА сosB = RHS =

LHs _ сos(А =

+

+

sinА cosB - сosА sinB(1 &

+

сosА cosB

=2cosA cosB

sinА sinB(3

RHS

=

сos2А ЦQ lv _- сosА+sinА

+

= sin 3А

=

=

LHs

sin(2А

+

I

А)

sin2А сosА + cos2A sinА(l) = 2 sinА сosА x сosА + (1 -2 sin2А) x sinА(7 & = 2 sinА сos2А + sinА -2 sin3А *2 sinзА = 2 sinА (1 -sin2А) + sinА

=3sinА-4sin3А=RHS sinА . сosА

Lг|D =БьsА_"ъjГд..

сos2А

=- sinА сosА 12 = =

Бi;7ъъъ7

=

йс/)

RHS

=

-]iБfi;oФ-_ sin(O

11)

1s0 -15o' -з0' 30' 150 -148.28' -58'28' 31'72'122.72 -165, -105, 15,75 -157.5, -57.5, 22.5, 112.5 -150, -30,30, 150

-120,0,120

0, +60, +120, +180

9 r30,190,1150 10 -157.5, -67.5,22.5,

112.5

Exerсise 11.l (page 114)

45)' V20 sin(x + 21 .80)" V1Osin(x + 288.43)" т Фcos(х+f,п)

1 3 5

9

10

гnZъ;;72

sin0сosФ+sinФcos0

=?ffi4o)=RHS

& 4)

4 5 6 7

=

+

сosQ

+

з

sinА)(сosА - sinА) сosА + sinА coSА - sinА = RHS

sin2А

RHS

ф) =F*фй7(1)

2

,^'

сosА+sinА +

=

'r -90,30,90,

сos2А - sin2А

(сosА

i0

= "jng *' sinф сosф

= 2)

сos10

---!sin

Еxerсise 10.4 (page l09)

|Ll

--'-I

7 LHs

B) + сos(А * B)

CoSА сosB - sinА sinB

=-1=*сos0 sin 0 sin 0

d2sin8

^'[2sin(х

+

-|п)

2 4 6 s

13 sin(x + 67.38)'

{2 sin(x + 135)"

{Тбsin(x + 108.43)"

l2cos(х-\п)

Both funсtions have the form R сos(x + с) with в = VsO. This means, see Seсtion 1.t.3, that in both сases the maximum value is Vso.

Еxerсise l1.2 (page l17)

't

з 5 7 9

11

1з

2

0,90,360

4

40.21,252.41 1s9.24,283.13 -180, -90, 't80 -.69.20,32.33 289.59,349.20

6

I

't0 12

16 LHs 90,330 45.24,180

14 15 2atх =_15and 165'-2 atx=-105and75 16 12 at х = -22.5 and 157.5' _ц[2 at х = -112'5 and 67.5 17 7 atх = 36.87, -3 atx = -144'13 18 4.7З2 at x = -1 7.63 and 1 62'37' 1'268 at х = -1 07.63 and 7 2.37 l(sin40

+

z

sin20)

}

(sin

.10")

10

2 оosC

+

sin20")

оosB

сos(C

=

+

i

2 4 6

в

9

tan0 11 0,120,240,360

1

1з 14

0' 90' 120' 180' 240'27o,360

2

+

sinB)

z sin}И = z sin}И =

+ в1 +

+

sin(А

в)(сos}И - в1- сos}(А

+ 8))

I {Gсm em

-"'

RHS =bcosC+ссosB 28 sin8 сosO = 2Е(sinB оosO

=

+

2R sinO сosB

+

sinO сosB)

+

sinC)

C) =2ЯsinА=2=LHS

= 2R(sin(B +

2 sin3А sin2А 2 оos36. sin.t3. +

5

Р)

10 tan20 12 0,120,180,240,360

LHS

=g

=!r@+b+c) =

B(sinА

+

sinB

sinlA сos}я + (sinB + sinO)) + с1 сos](в = R(2 sin'Acos]я + z sin}(8 = zR(sinlAсos}я + сos }я сos,1 (8 - с))

= л(2

+

B)

сos,1(А - B) +

в)(сos}(А - в1

+

2sin}И cos}(А

=z cosf,C(2 сos}я сos}в1 = 4 сos}А сos}в сos}c = RHS

+

+

в)

в))

сoslИ

в) сos] (А

}А sin ,1в sin } C = RHS

--

45,135,180,225,315 = sinА + sinB + sinO (sinА

+

sin50"sin60'sin 70"

15 LHs

=

z sin}И

+

10сm

2 сosЗA siп2A

tant@

+

=zcos|C(2sin}я sin}в1 = + sin}А sin}B сos}C = RHS .|7 LHs CoSА + сosB + сosO - 1 = (сosC - 1) = (сosА + сosB) + _ B) - 2sin2]C = z cos|(А+ в) cos} И = z sinlC 1сos}(А - s) - sin}с) = z sinz1C 1сos,1И -в) - cos}(А + в)) = z sinlC (2 sin}А sin}в1

B) + cos(C. B)

Еxerc se 1 2.2 |page1 25| 1 2 sin3А сosА 3 -2 sin30 sinO 5 2 сos41" сos6" 7 сot15"

sinB - sinO

Exerсise l3.1 (pagel3O)

в0. - sin s0 - sin 20)

4 ; (sin + D)) D) + сos(C 6 }(sin90. - sin30.) = ] }(оos3(C 7 cos2A - сos4А в !(sinос - sin 10D) 9 2 сosO sinD = sin(C + D) - sin(C-D) }(сos80"

z

+

sin8) - sin(А + B) sinf,(A+ в) сos}(А - B) _ 2sin}(А

= + sin

Еxerсise 12.1 (pagel2l}

1 3 5

=

(sinА

=

-'180,60, 180 -157.38, -90 -135, -75,45, 105

.1

at

sinА

=

45,90,225,270

x = 16.57, _t/s at x = - 63.43 5 at х = -53.13, - 5 atx = 126.87

{5

=

+

B)

zB сos}А(sinf,A + сos}(в _ с) _ с)) = zЕ сos}А(сos}{в + с1 + сos}(в =zн cos|А(2 сos}в сos}с) = 4R cosьA сos]в сos}с = RHS

=

с))

+

B)

5

Еxerсise 13.2 (page 137) f r/lsоm2, }{lsсm, f VGсm,,1{isсm,!{Gсm

t

6{з z^ _=-сm 3

7+

LHS

IabsinC =--

=m

i@+b-c) 2RsinАx2Psin8хsinC

{з(7 _{1з)

ц!13

6

Using Еquations 3 and 4:

LHs

=

ffд|вtc

Ah sinАsinBsinC =.rтffiБЪ

Ц=o,=RHS

.,' --DEt

AAAA s s-a ^_---; s_o s-с

2sin z}Асos.uл x zsin,18 cos}8 x zsin } Cсos}C ' + B|сos t И - B) -2 sin } о сos} о

.-

N

4 LHS=

=Гс ^

lв -2€tnlи.сoslАsin z z

A s }аbsinC

вsin 1А ^^ - - z

!(a+b+c)

^^

2

2RsinA+2Rsin8+2RsinC =

<

сostИ -в)-сos21И

+-ъiБЕ_;ъ;n"

=

-

x 2sin}B сos}в x 2sin}C сos}с 2sin|A сos}я / Z

-

вsin }А сos }я sin } всos }вsin,1 ссos z сos } C(сos Ltа - вl + sin } C) sin сos'1я sin,1 в сos} вsin } с

2sinlИ

+

в) сos,1И - B)

^^zz

-

-

-oDЩ

"1А

сos}И-B)+сosi$+B)

=8R =

4вsin}Аsin}вsin}c

+

RHS

B)

4Есos}Асos,1вsin}о

=

RHS

2sin}C сos]с }

с

RHS

=

=

= =

+

=8Е

sinАsinBsinC

^^

сos ] C(сos Ьtд - вl - sin

}

sin}всos}вsin }с z Z - sin]Асos}я

2ЕsinАx2RsinBxsinC

^h Zг1ъinА

сos}вsin] z Ccos}с C z

ff"ot2л

2t sinB - 2B sinO сos}я

'ьъi;Б*zяъinсi^ sinB _ sinC сos}А sinB

+

sinC

..

*г

sin]я

z сos - c)', сos }1! =@^"inlТ }

(B + с1 sin

}

-x-j

(B

!я sin}tв - C) .- cos}z\A I сosЁtлcos|tB_C).' sin}А sin

sinl(B - c)

=;щ

_ tan}{в - C) = LHS

о)

t'*l tl

I RHS =(s-a)x

tl

э

a.

_ r1"-\БG-aXs-bDF-с) s A

o

s

х

=r=LHS

Еxerсise 14 (page 1Ц|

ln all the answers to this exerсise, л and m are integers. |t is possible that your answer might take a different form from the one given and still be оorreсt.

l 3 5 7 9

180л + (_1),60

180л-45 120л + 60 18On

180mor90m+45

11 |nп - $Fl)"lт 1з |nп-}п 15 f,nп 17 lnп +\п 19 tnп

2

4

360л + 60 180m or 180m + 120

6 60n-30

I

1O

72n (4m + 1)45 or

12 |nп !f,п

fi

16

1s 2o

|mп orltzm |nп + f;п

$m

+ 1)п

ecirсles 132

ambiguous сase 68 angle of depression 145

equation equation equation equation equation equation

anglr of elevation 1tl5 angle offoсus 14

angle of slopе 28 area of triangle 63, 134 Aryаbhata | 4

+ 1)221

asinx+ 0сosx

-f,п

bearing 145 box problems 39

_ iЪ(4л + 1)п or |{+n l|lт 2'nп

adjaсent 14

cosx= c 57 CoSX= сosсr 141 sinx= с 54 sinx= sinсr 139

tаnx= с 59 tаnx= tanсr 141 equations 92,108

111

сa|сulator, use of 27

сhords 3

faсtor formu|ae 119 faсtorformuIae, second set 122 general solution of equations 139

gradient 28

сirсular arс, |ength of 8.4 сirсular seсtor, area of 87 сirсumсirсle 127 сompound angle 97 сorresponding ang|es 145 cos45o, оos30o, cos60o 25,26

height of distant objeсt 73

сos(,4

il|ogiсa| method of proof 108 inсirсle 131 inverse сosinе 21 inverse sine 21

fl 99 - 0 100

cos(t+

сosecaпt 91 оosine 19,20,46 сosine formuIa 69

Heron's

formula 134

Hipparсhus 3 horizontal distanсe 75

hypotenuse 15 identities 106

оotangent 91

inverse tangent 9

degree, radian сonversion 86 distanсe, inaссessib|е objесt 75

Joseph, Geоrge

Gheverghese 3

seсant 91 similar triangles 6, 147

kite 146 mode, of оalсu|аtor 9 multiple angle formulae

lffl

notation for sides аnd ang|es 6i}

opposite 14 pеriod 48,52j8 periodiс 49' 146 prinсipa|ang|e 54 projeсtions 28 Ptolemy 4 pyramid рroblems 36 Pythagoras's theorеm

Regiomontanus 4 Яsin(O + а| 111

0 sin(t- 0 S

siп60. 25

sine 19,20,46

sine formula 66, 128 sIant height of a сone 147

slope 28 survеying 73 tап 45o, tan 30o, tan

tan(/ +

O

tan(t- O

9',1ф

radian, degree сonversion 86

radiаns 84

sin45o, sin30o, 97 sin(/ +

100 100

tаngent 8,5{l Thа|es 1' 6 triangulation 76 trigonometry 3 Varahаmihara 4 wedgе prob|ems 41

60o 25, 26

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Are you new to trigonometry? Do you need praсtiсe for an exam or сourse? Do you need to refresh your understanding?

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