Ch6 - Trigonometry - Prelim Maths

7/25/2019 Ch6 - Trigonometry - Prelim Maths

1/41

Trigonometry Chapter 6

Trigonometric Ratios and the Calculator

Angles can be given in degrees, minutes and seconds. We usually use angles in

degrees and minutes.

25 32 '43 ' '

We need to know how to round of angles correctly. ust like time, minutes and

seconds are out o! 6", not #"", i.e.$

To %nd unknown sides in right&angled triangles, you will use the !ollowing buttons

on your calculator$

sin cos tan

To calculate angles you will use$

sin1

cos1

tan1

To change a decimal to degrees, minutes and seconds, or vice versa, use the

button$

' '( or )*+

Examples

#. ound of to the nearest

degree$-.

a 23 16 '34 ' '

/.

b 52 43 '14 ' '

0.1. ound of to the nearest

minute$6.

c 23 16 '34 ' '

2.

d 52 43 '14 ' '

3.

4. Change 34 54 ' to a

decimal.

#".

##.Change 76.42 to degrees

and minutes.#-.#/.5ind correct to - decimal

places$#0.

a sin12 38 '

#1.

b tan 56

13

'#6.

age 7 #

6" minutes 8 9999 degree

(60'=1 )


2/41

#2.5ind in degrees and

minutes$#3.

a sin=0.456

#4.

b cos =( 35 )

age 7 -


3/41

20.

21. Answers

#. ound of to the nearest degree$--.

a 23 16'34

' '=23

-/.

b 52 43'14

' '=53

-0.-. ound of to the nearest minute$-1.

a 23 16 '34 ' '

-6.

b 52 43 '14 ' '

27.

/. 34 54'=34.9

-3.-4.

0. 76.42 =76 25'12

''

/". 7625 '

/#.1. 5ind correct to - decimal places$/-.

a sin 12 38'=0.21871

//. 0.22

/0.

b tan 56 13'=1.4947

/1. 1.49

/6.

6. 5ind

in degrees and minutes$/2.

a sin=0.456

/3. = sin1

0.456

/4. 27.129

0". 27 7'46 ' '

0#. 27 8'

0-.

b cos=( 35 )

The minutes are less than /"

so you round downto the

The minutes are more than/" so you round upto the

The seconds are more than

/" so you round upto the

The seconds are less than /"

so you round downto the

Type in /0 : ' '( 10 : ' '( enter ;or

8.

: ' '(

Type in 26.0- then : ' '( . Write

what your calculator says and

then round of to the nearest

Type in sin then #- : ' '( /3 : ' '(

then to close the brackets and

Type in tan then 16 : ' '( #/ : ' '(

then to close the brackets and

Type in shi!t ;or -nd sin then

".016 then to close the brackets

and then enter ;or 8. ress : ' '(

and then round appropriately.

Type in shi!t ;or -nd cos then

35 then to close the

brackets and then enter ;or 8.

: ' '(


4/41

hypotenuse

ad=acent

ad=ace

nt

hypotenuse

>

>

0/. = cos

1( 35 )00. 53.1301

01. 53 7'48.368 ' '

06. 53 8 '

02.

48.

49. Trigonometric Ratios

1".To calculate the length o! a side or the si?e o! an angle in right&angled

triangles, we can use certain ratios.


5/41

66.

62.

63.

64.

2".

2#.

2-.

!. Reci"rocal #dentities

20.TAB $ # 8 +C $ # 8 CD+C $ # 8 CDT $ # 8

21.

26.TAB 8 +C 8 CD+C 8CDT 8

22.

23.Eere are the / reciprocalratios or reciprocal identities$

79.

80.

81.

82.

83.

84.

8.

8!.

32.

88.

89.

90.

91.

92. Examples

#. Write down the ratios o! sin ,cos , tan , cosec , seccot .

Cosecant cosec = 1

sin =

Secant sec= 1

cos =

Cotangent cot = 1

tan=


6/41

#2

#1

3

>

#2

#1

3

>

Dpposite ;D

Ad=acent ;A

Eypotenuse ;E

>

4/.40.41.46.42.43.44.

-.

sin=2

5 , %nd the eact ratios o!

cos , tan,cosec,seccot

.

100. Answers

#.#"#.

#"-.

#"/.

#"0.

#"1.sin =

O

H=

8

17

cosec = 1

sin =

1

8

17

=17

8

#"6.cos

=

A

H=

15

17

sec= 1

cos =

1

15

17

=17

15

#"2.tan =

O

A=

8

15

cot = 1tan

= 18

15

= 158

-. sin=

2

5

#"3.#"4.##".###.

##-.##/.##0.##1.##6.

Food idea to label

sides. Dnce you

become !amiliar

they should become

To begin, draw a right angled

triangle with an unknown

angle. Bet use the !act that

sin =O

Hand write in these

sides on the triangle. To %nd

the third side use tha oras(

1 -

sin = OH

= 25

cosec = 1

sin =

1

2

5

=5

2


7/41

>

4" >

;Gsing the angle sum o! a triangle is #3".A

HCa

b

c

##2. c2

=a2

+ b2

##3. 52

=a2

+ 22

##4. 25=a2

+4

#-". 44

#-#. 21=

a

2

#--. a=21

#-/.#-0.

#-1. Co$unction #dentities

#-6. Gse your calculator to evaluate the !ollowing to /decimal places

#-2.

#-3. +in -": 8 cos 2": 8

#-4.

#/". +in 1": 8 cos 0": 8

#/#.

#/-. +in /1: 8 cos 11: 8

#//. What have you noticedI

#/0.

#/1. Why do you think the answers are the sameI

#/6.

#/2. We have established our %rst important result, namely that

138. " the sine o# an an$le is e%ual to the cosine o# it&s

complement "

#/4. This is represented using the mathematical symbols

140. 'in ( ) cos *90 + ( ,

%4%.

#0-.

#0/.

#00.

#01.

#06.


8/41

#02.(90 ) =sin=sin

#03.(90 )=cos=cos

#04.(90 ) =tan=tan

#1".sec =

1

cos=

1

a

c

=c

asec (90 )= 1

cos (90 )=

1

b

c

=c

b

#1#.

(90 ) = 1

sin (90 )=

1

a

c

=

cosec = 1

sin =

1

b

c

=cosec

#1-.

(90 ) = 1tan (90 )

=1

a

b

=

cot = 1

tan =

1

b

a

=cot

#1/. 5rom these ratios come the

results$

#10.

#11.

#16.

#12.

#13.

#14.

#6".

#6#.

#6-.

sin = (90 )

cos= (90 )

tan = (90 )

cot = (90 )

sec= (90 )

cosec = (90 )


9/41

#6/.

#60.

1!.

1!!. Examples

#. +how sin 32 =cos58 .

#62.

-. +impli!ysec 65

sec 65 + cosec25

#63.

/. 5ind the value o! x i! tan 27=cot (x+20 ) .

#64.

0. 5ind the value o! p i! sin (2p+15 )=cos (3p20)

170. Answers

#. sin 32=cos (9032 )

#2#. cos68

#2-.

-.sec 65

sec 65 + cosec25 =

sec 65

sec 65 +sec (9025 )

#2/. sec65

sec 65 + sec 65

#20.

sec65

2 sec 65

#21.

1

2

#26.#22.#23.

/. tan 27= cot (x +20)

#24.

#3". tan 27=cot (9027 )

#3#. cot 63

#3-.

#3/. cot (x+20 ) =cot 63

#30.

#31. x +20=63

#36. 2020

#32. x=43

#33.

+ince - o! the / ratios are

sec change the third one to

sec. Ji.e.

+impli!y by collecting like

terms on the bottom.

+impli!y by cancelling

sec65 !rom to and

These should now be the

same because they are

These should now be the

same because they are both


10/41

#- cm cm

#/.0 m

y m

a b

0. sin (2p+15 )=cos (3p20)

#34.

#4". sin (2p+15 )= cos [ 90 (2p +15 )]

#4#. cos [902p15 ]

#4-. cos (752p )

#4/.

#40. cos (3p20 ) =cos (752p )

#41.

#46. 3p20=752p

#42. +2p+ 2p

#43. 5p20=75

#44. +20+20

-"". 5p=95

-"#. 5 5

-"-. p=19

-"/.

&'4.

&'(.&'). Right-Angled Triangle Pro*lems

-"2. Trigonometry can be used to %nd unknown sides or angles in

triangles.

208. -indin$ a 'ide

-"4.

;other than the right angle you can %nd the length o! any other side in thetriangle using the trigonometric ratios. *ake sure you use the ratios

correctly, as the unknown side can be on the top or the bottom o! the

!raction when you try and solve.

210. Examples

#. 5ind the length o! the unknown side to - decimal places in the !ollowing

triangles$-##.-#-.

-#/.-#0.-#1.-#6.-#2.

He care!ul when taking away.

Always put in brackets and

then multiply out. He care!ul


11/41

-#3.

219.

220.

221. Answers

#. Locate the known angle, decide what sides you have and then which ratio

to use$---.

a < have an opposite side and a hypotenuse so it is sine.

--/. sin=

O

H

--0. sin 23 45

'=

x

12

--1. 1212

--6. 12 sin 23 45'=x

--2. 4.8329 =x

--3. x=4.83

--4.b < have an ad=acent and hypotenuse so it is cosine.

-/". cos=

A

H

-/#. cos67 12'=13.4

y

-/-. y y

-//. y cos67 12'=13.4

-/0. cos67 12' cos67 12

'

-/1. y=34.57925 .

-/6. x=34.58

-/2.

238. -indin$ an An$le

-/4.

%nd the si?e o! any o! the other two angles in the triangle using the

trigonometric ratios. emember to %nd angles we will be using

sin1

,cos1

, tan1

.

240.

241. Examples

Botice how the pronumeral is on the

bottom. He care!ul and don(t mi themround as you will get a diferent answer.

+ome people like to always put the

pronumeral on the top o! the !raction.


12/41

a b3.- cm

#".2 cm

16 m

2- m

#. 5ind the si?e o! the unknown angle to the nearest minute in the !ollowing

triangles$

-0-.

-0/.

-00.

-01.

24!.

247.

248.

249.

20. Answers

#. Locate the unknown angle, decide what sides you have and then which

ratio to use$-1#.

a < have an ad=acent side and an hypotenuse, so it is cosine.

-1-. cos=

A

H

-1/. cos=

8.2

10.7

-10. = cos

1( 8.210.7 )-11. =39.9722

-16. =39 58'20.271 ' '

-12. =39 58'

-13.-. < have an opposite side and an ad=acent side, so it is tangent.

-14. tan=O

A

-6". tan =

56

72

-6#. = tan

1( 5672 )-6-. =37.8749

-6/. =37

52

'29.941

' '

-60. =39 52'

-61.

emember to close the

emember i! the seconds are

less than /" you round down

to the nearest minute.


13/41

-66.

&).

&)8.


14/41

&)9. A""lications

-2". Trigonometry can have many practical applications, !rom such areas

as building, construction, surveying and navigation.


15/41

278. earin$s

-24. Hearings are based on directions related to the compass. True

*earingsmeasure angles cloc+,ise!rom orth. Hearings are usually

written as / numbers e.g. 124 ,048 , etc.

-3". Always draw a diagram i! one is not provided.-3#. The bearing o! a point is the number o! degrees in the angle

measured in a clockwise direction !rom the north line to the line =oining the

centre o! the compass with the point.

-3-. A bearing is used to represent the direction o! one point relative to

another point.

-3/.

-30. 5or eample, the bearing o!A !rom Bis "61M. The bearing o! B !romAis -01M.

285.

-36.

287. Examples

#. 5rom a pointA, level with the !oot o! a vertical pole and 25m !rom it, the

angle o! elevation o! the top o! the pole is 40 . Calculate$

a The height o! the pole ;to the nearest metre,b The angle o! elevation !romAo! a point hal!way up the

pole ;to the nearest minute.-33.

-. An observer in a lighthouse 100 m above sea level is watching a ship

sailing towards the lighthouse. The angle o! depression o! the ship !rom

the observer is 15 .

a Eow !ar is the ship !rom the lighthouseI Correct to the

nearest metre.


16/41

b +ome time later, the angle o! depression is measured as

25 . Eow !ar has the ship travelledI Correct to the

nearest metre.-34.

/. Bick cycles 15km due north, then 12km due east and %nally 20km

due south. What are his distance and bearing !rom his original positionI

-4".

291.


17/41

A

292. Answers

#. )raw a diagram$-4/.-40.-41.-46.-42.

-43.-44./"".

/"#. tan 40 =

h

25tan =

10.5

25

/"-. 2525 =tan1( 10.525)

/"/. 20.97749=h 22.782

/"0. h=21m 22 47 '

/"1./"6./"2.-. )raw a diagram$/"3./"4./#"./##./#-./#/.

/#0./#1.

/#6.

/#2. tan 15=

100

x

tan 25=100

y

/#3. x x y y

/#4. x tan 15 =100y tan 25=100

/-". tan 15 tan 15 tan 25 tan 25

/-#. x=

100

tan 15 y=

100

tan 25

/--. 373.2050 =214.450

/-/. 373m 214m

/-0.

DistanceTravelle=373214

/-1.159m

b

a

21 mh

10.5 m

25m25 m

A

Bow that you

have !ound the

height you can

use it in part b

to help solve,

but remember

to halve it !or

'hal!way up theole(.

ba25

Gse alternate

100 m

25

yx

)on(t !orget to

answer the @uestion.

-#0 m is how !ar it is

still !rom the

lighthouse, we want

to know how !ar it


18/41

32!.

327.

328.

329.

330.

/.

331.

332.

333.

334.

33.

33!.

337.

!!8.

12m

Gsing =ust the triangle the

height is 5 m , i.e.15m

tan =5

12

= tan1( 512 )

22.6198

'

20 m

5m


19/41

#

#

--

#

!!9. /act Ratios

/0". We can %nd the eact values o! certain angles, that being

30 ,45 60 . o! # we can %nd the hypotenuse using ythagoras(s Theorem$

/0-.

/0/.

/00.

/01.

346.

347.

/03. Gsing the above triangle we get the eact ratios$

/04.

/1".

351.

352.

353.

/10. We can %nd the ratios !or 30 and 60 using an e@uilateral

triangle and then splitting it in hal!. 5or this we use sides o! - so that when

we split in hal! we can have a side o! #. +imilar to above, we can %nd the

height ;a short side by using ythagoras( Theorem$

/11.

/16.

/12.

/13.

/14.

360.

/6#. Gsing the above triangle we get the eact ratios$

/6-.

c2

=a2

+ b2

12

+12

1+1

4

sin 45= 1

2

cos 45= 1

2

c2

=a2

+ b2

22

=a2

+12

4= a2

+ 1

sin 60=3

2

cos60 =1

2

sin 30 =1

2

cos30 =3

2

1


20/41

/6/.

/60.

365.

366.

3!7. Examples#. 5ind the eact value o! $

a sin45 +cos 45

b 3 sec 60

c sin2

30 +cos2

30

/63.

-. 5ind the eact value o! x $

/64./2"./2#./2-./2/.

374. Answers

#.

a sin 45+cos 45=

1

2+ 1

2

/21. 1+1

2

/26.

2

2

/22. 22

2

/23. 2

/24./3".

b 3 sec 60 =3

1

cos 60

/3#. 3

1

1

2

/3-. 3 1

2

1

/3/. 6

/30.

Nou can

rationalise the

denominator by

multiplying top


21/41

5irst Ouadrant+econd Ouadrant

hird Ouadrant5ourth Ouadrant

c sin230 +cos

230 =( 12 )

2

+(32)2

/31.1

4+3

4

/36. 1

/32.-.

/33.

/34. Alsotan 30=

1

3

/4".

x

12=

1

3

/4#. 12 12

/4-. x= 12

3

/4/.

!94. Angles o$ 0agnitude

/41. The angles in a right&angled triangle are always acute ;ecept the

right angle o! course, but we may need to know angles greater than

90 , such as with bearings. Also negatives angles are used in diferent

situations such as engineering and science. Angles are measured around a

circle starting !rom the positive direction o! thex&ais.

/46.

/42.

/43.

/44.

0"".

0"#.

0"-.

0"/.

404.

0"1. When %nding the value o! angles greater than 90 we always

begin at 0 and move in an anti&clockwise direction. The angle is

always taken of thex&ais ;i.e. the hori?ontal line. The sign o! the ratio

Nou can rationalise thedenominator and get


22/41

S

A

TC

S

A

T C

can be determined by the !ollowing acronym$ A+TC ;All +tations To

Central.

0"6.

0"2.

0"3.

0"4.

0#".

0##.

0#-.

0#/.

0#0.

0#1.

0#6.

0#2.

0#3.

0#4.

420. Examples

#. 5ind the eact value o!$

a sin135

b tan 300

c cos (150 )

d sin 690

421. Answers

#. 5or the answers to be eact you know you are going to use angles o!

30 , 45 60 .


23/41

S

A

T C

S

A

T C

0-2.

0-3.

b tan 300 =tan60

0-4. 3

0/".

0/#.

0/-.

0//.

0/0.

0/1.

c cos (150 )=cos30

0/6.

0/2.

0/3.

0/4.

00".

00#.

d

00-.

1

2

00/.

000.

001.

006.

002.

003.

449.

The angle it makes with the

x&ais is 60 and only cos

is positive in this @uadrant, so

tan has to be negative

tan 300 =tan 60

Hecause it is a negative

angle you =ust go backwards

;i.e. clockwise. The angle it

makes with thex&ais is

30 and only tan is

positive in this @uadrant, so

To get 690 go around the

number plane once and then

an etra 330 ;

690360 =330 . The

angle it makes with thex&ais

is 30 and onl cos is


24/41

S

A

T C

S

A

T C

4('. Trigonometric 1uations

451. A trigonometric e@uation involves %nding an angle whose

trigonometric ratio is given. There is usually more than one solution to an

e@uation. Also be care!ul to see what the conditions o! the e@uation are

e.g. 0 ! !360180 ! !180 etc. This eercise is the opposite to the

last eercise, we are working backwards. Dnce again it is a good idea todraw diagrams !or these @uestions.

42. Examples

#. +olve !or 0 ! !360 $

a sin=

1

2

b tan =3

c 2cos2=1

01/.

-. +olves

2=

1

2 !or180 ! !180 .

44. Answers

#.

a sin=

1

2

4.

4!.

47.

458.=30 ,18030

014. 30 ,150

4!0.

b tan =3

06#.

06-.

06/.

060.

sin 30 =1

2 and it is positive.

+in is positive in the %rst and

second @uadrants. We make -

angles that are 30 !rom thex&

ais in these @uadrants.

tan 60=3 , but it is negative.

Tan is negative in the second and!ourth @uadrants. We make -

angles that are 60 !rom thex&

ais in these @uadrants.


25/41

A

T C

465. =18060 ,36060

066. 120 , 300

062.

063.

064.

02".

c 2cos2=1

02#. 22

02-.

02/.

020.021.026.022.023.024.03".03#.03-.03/.030.031.

486. 2=60 ,36060 ,360 +60 ,72060

487. 60 ,300 ,420 ,660

033. =30 , 150 , 210 , 330

034.04".

04#.04-.04/.040.041.046.

-. +olves

2=

1

2 !or18 0 ! ! 180 .

042. sin =" 12043.

" 1

2

Can(t get an eact value with # using cos, so

divide by - and then we can use Q which has an

cos60 =1

2 , and it is positive.

Cos is positive in the %rst and

!ourth @uadrants. We make -

The conditions !or this @uestion are

0 ! !360 , but we have 2 , so

multiplying everything by - we get the

condition 0 ! 2 ! 720 , so we are going

)ividing everything by - so

we et instead o! 2 .

Botice how the 0 answers are

within the original conditions

o! the @uestion$


26/41

499.

00.

01.

02.

03.

504. =180 +45 ,0 45 ,45 ,180 45

1"1. 135 ,45 , 45 , 135

(').

To get rid o! the s@uared we s@uare

root. Hut remember when we s@uare

root in an e uation we alwa s et " .

sin 45= 1

2 , and it is positive and

negative. +o we make 0 angles that45 &


27/41

('. Trigonometric 2ra"hs

1"3.


28/41

1#/.

1#0. The height o! the graph ranges !rom to #. The graph is

symmetrical and continues with the same shape a!ter 360 .

1#1.

1#6.

1#2. y=tanx

1#3.

519. There are asymptotes at x=90 270 . The graph continues with

the same curve beyond 360 .

20.

#


29/41

1-#. y=cosec x

1--.

1-/. This graph is the inverse o! y=sinx . Hecause you can(t divide by

?ero, we have asymptotes at x=180360 .

1-0.

525.

1-6. y=secx

1-2.

528. This graph is the inverse o! y=cosx . Hecause you can(t divide by

?ero, we have asymptotes at x=90 270 .

#


30/41

1-4. y=cotx

1/".

531. This graph is the inverse o! y=tanx . Hecause you can(t divide by

?ero, we have asymptotes at x=180360 .

532. 5rom all these graphs you can deduce what the values o!

90 ,180 ,270360 .

(!!.


31/41

(!4. Trigonometric #dentities

1/1. There are a number o! trigonometric identities that can be used to

simpli!y @uestions.

manipulate to get the identities that are related to it.

1/6.

1/2.

1/3. cot =

1

tan

1/4.

1

sin

cos

10".

cos

sin

10#.10-.10/.100.101.106.102.103.104.11".

11#.

11-. Hy dividing by sin2

:

11/.sin

2

sin2+cos

2

sin2=

1

sin2

110. 1+cot2

= cosec2

111. Hy dividing by cos2

:

116.sin

2

cos2

+

cos2

cos2

=

1

cos2

112. tan2

+1= sec2

113. we get the - identities$

114.

16".

RR tan =

sin

cos

cot=cos

sin

RR sin

2+cos

2=1

1+cot2=cosec

2

tan2

+1= sec2


32/41

16#.

16-.

!3. Examples

#. rove the !ollowing identities$160.

a1sin

2

sin

2+cos

2

=cos2

161.

b tanA sinA +cosA =sec A

166.

c sin2

tan +cos2

cot +2sin cos =tan +cot

!7. Answers

#. rove the !ollowing identities$163.

a1sin

2

sin2

+cos2

=cos

2

164.12".

12#. #H$= 1sin

2

sin2+cos

2

12-. cos

2

1

12/. cos2

120. %H$

121.

b tanA sinA +cosA =sec A

126.122. #H$=tanA sinA +cosA

123.

sinA

cosA sinA+ cosA

124. sin

2A

cosA +

cosA

1

13". sin

2A

cosA +

cosA

1( cosAcosA)

13#. sin 2A

cosA +

cos2A

cosA

13-.

+tart with the side that you think you

can manipulate. Gsually it will be the

Le!t Eand +ide ;LE+. Food idea to

sin2

+cos2

=1

sin2

sin2

2 = 2

Another good

idea is to write to

the side o! your

working what

Dnce it is e@ual

to the ight

Eand +ide ;E+

you %nish by

tanA =sinA

cosA

Beed the same

denominator to add

!ractions, so

multiply top and


33/41

13/.

1

cosA

130. sec A

131. %H$

136.

c132.

133. #H$=sin2

tan +cos2

cot + 2sin cos

134. sin

2

sin

cos + cos

2

cos

sin+ 2sin cos

14". (1cos2 ) sin

cos +( 1sin2 ) cos

sin +2sin cos

14#.

1sin

cos cos

2

sin

cos+1

cos

sinsin

2

cos

sin + 2sin cos

14-.

sin

cos sin cos +

cos

sin sin cos + 2sin cos

14/.

sin

cos +

cos

sin

140. tan +cot

141. %H$

146.

142.

143.

99.

sin2

+cos2=1

sin2

sin2

5rom the %rst to the second

line o! working < used the

identities$

tan =sin

cos cot =

cos

sin


34/41

A

H Ca

bc

#-.#

#3./

)''. on-Right-Angled Triangle

Results

6"#. A non&right&angled triangle is named so that its angles and opposite

sides have the same pronumeral ;i.e. angle A is opposite side a, etc.

6"-.

6"/.

6"0.

)'(.

6"6.

!07. he 'ine ule

6"3. 5or %nding a side$

6"4.

6#".

6##. 5or an angle we =ust Sip it upside down$

6#-.

6#/.

6#0.

6#1. Nou can use this rule in non&right&angled triangles when you have -

pairs o! opposite sides and angles.

!1!. Examples

#. 5ind the length o! x in the !ollowing to # decimal place$

6#2.

6#3.

6#4.

6-".

-. 5ind the si?e o! to the nearest minute$

6-#.6--.6-/.

6-0.6-1.6-6.6-2.

a

sinA=

b

sin&=

c

sin

sinA

a =

sin&

b =

sin

c


35/41

#-.#

#3./

!28. Answers

#. Gsing the sine rule$

6-4.a

sinA=

b

sin&=

c

sin

6/". xsin59 34 '= 12.1

sin 3212'

6/#. sin 5934 ' sin 5934 '

6/-.x=

12.1

sin32 12' sin 59 34

'

6//. 19.5783

6/0. 19.6 units

6/1.-.

6/6.sinA

a =

sin&

b =

sin

c

6/2. sin27.4= sin 28 8

'

18.3

6/3. 27.4 27.4

6/4. sin =sin 28 8

'

18.327.4

60". = sin1(sin 28 8

'

18.327.4)

60#. 44.910..

60-. 44 54'37.164

' '

60/. 44 55 '

!44.

Hetter to type this in your

calculator so you are not

rounding of until the last


36/41

A

H Ca

bc

/

2

#-

#"

6

/

2

!4. he osine ule

606.

602.

603.

604.

61".

61#. 5or a side$ 5or an angle$

61-.

61/.

610. Gse this when you have - sides and the angle in between. Gse

this rule when you have all / sides.

!. Examples

#. 5ind the length o! y in the !ollowing correct to - decimal places$

616.612.613.614.66".66#.

66-.66/.

-. 5ind the si?e o! ( in the !ollowing correct to the nearest minute$

660.

661.

666.

662.

663.

!!9. Answers

#. c2

=a2

+ b2

2 ab cos

62". y2

=32

+72

2 3 7 cos34

62#. 23.180421

62-. y=23.180421..

62/. 4.81460

620. 4.8 units

621.

cos=a

2+b

2c

2

2abc2=a

2+b

22abcos

)on(t round of until the %nalemember to s@uare root to

get rid o! the s@uared.


37/41

#-

#"

6

626.622.623.

-. cos=a

2+b

2c

2

2ab

624. cos(=10

2+12

26

2

210 12

63".

208

240

63#. (=cos

1( 208240 )63-. 29.9264

63/. 29 55'35.166

''

630. 29 56 '

631.636.632.

!88.

emember to %nd

=ust the angle use

Nou can but this

straight in your

calculator butremember to put (=cos1

[(102+12262)

(21012 )]


38/41

C

/" km

-" km

!89. 'ine and osine ro5lems

64". This eercise looks at using the sine and cosine rules in real li!e

situations.

!91. Examples

#. A, H and C are three towns such that H is -" km !rom A on a bearing o!330 and C is /" km !rom A on a bearing o! 204 . 5ind the distance

!rom H to C to the nearest kilometre.

-. An aircra!t Sies !rom a point A to a point H 0"" km on a course o! 040 .


39/41

2#". &A=180 60 43 51'

2##. 769 '

2#-.

a&

sin76 9'

= 500

sin60

2#/. sin 76 9'

sin 76 9'

2#0. &=

500

sin 60 sin 76 9

'

2#1. 560.5642434

2#6. 561km ;nearest km

b &earin)=040 +76 9'

2#2. 116 9 '

718.

Couldn(t use sine rule or cosine rule

yet to %nd the distance, but i! we

!ound a missing angle, i.e. HAC, wecould use the sine rule to %nd the

distance between A and H. To %nd

HAC we had to use the sine rule to

%nd HCA and then the angle sum o!

a triangle. To %nd the bearing < add

the an le HCA and the bearin o!


40/41

A

H Ca

bc

A

H C

%9. Area2-". We can %nd the area o! a triangle with a right angle using the

!ormulaA=

1

2bh

. We can also %nd the area o! a non&right angled

triangle using a similar !ormula$

2-#.

2--.

2-/.

2-0.

2-1. emember when we are %nding area we use mm2

, cm2

, m2

etc, or

*nits2

i! the units are not stated.

72!. Examples

#.


41/41

2/2. 58 45 '

2/3.

2/4. A=

1

2ab sin

20".

1

280 100 sin 58 45

'

20#. 3419.647

20-. 3420m2

;nearest m2

20/. nd o! Trigonometry

Chapter 6

To %nd the area you need an

angle so use the cosine rule

to %nd any angle in the

Ch6 - Trigonometry - Prelim Maths

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Transcript of Ch6 - Trigonometry - Prelim Maths