Mathematics MYP5+ (2nd edn) - 2010/12/3 9:16 - page 202

Chapter 10TRIGONOMETRY

EXERCISE 10A.1

1 a i sin µ =OPP

HYP=

p

r

ii cos µ =ADJ

HYP=

q

r

iii tan µ =OPP

ADJ=

p

q

iv sinÁ =OPP

HYP=

q

r

v cosÁ =ADJ

HYP=

p

r

vi tanÁ =OPP

ADJ=

q

p

b i sin µ =OPP

HYP=

y

x

ii cos µ =ADJ

HYP=

z

x

iii tan µ =OPP

ADJ=

y

z

iv sinÁ =OPP

HYP=

z

x

v cosÁ =ADJ

HYP=

y

x

vi tanÁ =OPP

ADJ=

z

y

c i sin µ =OPP

HYP=

4

5

ii cos µ =ADJ

HYP=

3

5

iii tan µ =OPP

ADJ=

4

3

iv sinÁ =OPP

HYP=

3

5

v cosÁ =ADJ

HYP=

4

5

vi tanÁ =OPP

ADJ=

3

4

d Let the unknown side be x.

) x2 + 42 = 72 fPythagorasg) x2 + 16 = 49

) x2 = 33

) x =p33 fas x > 0g

i sin µ =OPP

HYP=

4

7

ii cos µ =ADJ

HYP=

p33

7

iii tan µ =OPP

ADJ=

4p33

iv sinÁ =OPP

HYP=

p33

7

v cosÁ =ADJ

HYP=

4

7

vi tanÁ =OPP

ADJ=

p33

4

e Let the unknown side be x.

) x2 = 32 + 52 fPythagorasg) x2 = 9 + 25

) x2 = 34

) x =p34 fas x > 0g

i sin µ =OPP

HYP=

5p34

ii cos µ =ADJ

HYP=

3p34

iii tan µ =OPP

ADJ=

5

3

iv sinÁ =OPP

HYP=

3p34

v cosÁ =ADJ

HYP=

5p34

vi tanÁ =OPP

ADJ=

3

5

f Let the unknown side be x.

) x2 = 42 + 72 fPythagorasg) x2 = 16 + 49

) x2 = 65

) x =p65 fas x > 0g

i sin µ =OPP

HYP=

7p65

ii cos µ =ADJ

HYP=

4p65

iii tan µ =OPP

ADJ=

7

4

iv sinÁ =OPP

HYP=

4p65

v cosÁ =ADJ

HYP=

7p65

vi tanÁ =OPP

ADJ=

4

7

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2 a sin µ =OPP

HYP

) sin 70± =x

a

b sin µ =OPP

HYP

) sin 35± =x

b

c tan µ =OPP

ADJ

) tan 64± =x

c

d cos µ =ADJ

HYP

) cos 40± =d

x

e cos µ =ADJ

HYP

) cos 49± =x

e

f tan µ =OPP

ADJ

) tan 73± =f

x

g cos µ =ADJ

HYP

) cos 54± =g

x

h tan µ =OPP

ADJ

) tan 30± =h

x

i sin µ =OPP

HYP

) sin 68± =i

x

3 a sin µ =OPP

HYP

) sin 76± =x

16

) x = 16 sin 76±

) x ¼ 15:52

b cos µ =ADJ

HYP

) cos 30± =x

15

) x = 15 cos 30±

) x ¼ 12:99

c tan µ =OPP

ADJ

) tan 64± =x

4:8

) x = 4:8 tan 64±

) x ¼ 9:84

68°x

i

HYP

OPP

ADJ

30°

x hOPP

HYP

ADJ

70°

x

a

OPP

HYP

ADJ

35°x

b OPPHYP

ADJ

64°

x

cHYP

ADJ

OPP

40°x

d

HYP

OPP

ADJ

49°

x

e

ADJOPP

HYP

73°x

fOPP

HYP ADJ

54°x gHYP

ADJ

OPP

30°x cm

15 cmHYP

OPP

ADJ

76°

x cm

16 cm

OPP

HYP

ADJ

64°

x m

4.8 m HYP

OPP

ADJ

34°

x m8 m

HYP

OPPADJ

42°x km

5 kmOPP

ADJ

HYP

74°x cm

80 cm

HYP

OPP

ADJ

d cos µ =ADJ

HYP

) cos 42± =5

x

) x =5

cos 42±

) x ¼ 6:73

e tan µ =OPP

ADJ

) tan 34± =8

x

) x =8

tan 34±

) x ¼ 11:86

f tan µ =OPP

ADJ

) tan 74± =80

x

) x =80

tan 74±

) x ¼ 22:94

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 203

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g sin µ =OPP

HYP

) sin 28± =x

52

) x = 52 sin 28±

) x ¼ 24:41

h cos µ =ADJ

HYP

) cos 68± =x

45

) x = 45 cos 68±

) x ¼ 16:86

i tan µ =OPP

ADJ

) tan 27± =x

11

) x = 11 tan 27±

) x ¼ 5:60

j cos µ =ADJ

HYP

) cos 70± =5:6

x

k tan µ =OPP

ADJ

) tan 50± =27

x

l tan µ =OPP

ADJ

) tan 49± =12

x

4 a µ = 90¡ 28

) µ = 62

sin µ =OPP

HYP

) sin 62± =a

12

) a = 12 sin 62±

) a ¼ 10:60

cos µ =ADJ

HYP

) cos 62± =b

12

) b = 12 cos 62±

) b ¼ 5:63

b µ = 90¡ 63

) µ = 27

cos µ =ADJ

HYP

) cos 27± =15

a

) a ¼ 16:83

tan µ =OPP

ADJ

) tan 27± =b

15

) b = 15 tan 27±

) b ¼ 7:64

c µ = 90¡ 25

) µ = 65

sin µ =OPP

HYP

) sin 65± =45

a

tan µ =OPP

ADJ

) tan 65± =45

b

EXERCISE 10A.2

1 a tan µ =OPP

ADJ

=3

2

) µ = tan¡1( 3

2)

) µ ¼ 56:3±

b sin µ =OPP

HYP

=4

7

) µ = sin¡1( 4

7)

) µ ¼ 34:8±

28°x m

52 m

ADJ

HYP

OPP 68°x km

45 km ADJ

HYP

OPP

27°x mm

11 mm

OPP

HYP

ADJ

70°x cm

5.6 cm

HYP

OPPADJ

50°

x m27 mADJOPP

HYP

49°x km

12 km

HYP

OPP

ADJ

q°

28°b cm

12 cm

a cmOPP

ADJHYP

a mb m

15 m

63°

q°

OPPHYP

ADJ

25°a cm

b cm45 cm

q°

OPPADJ

HYP

3 cm

2 cm

qHYP

OPP

ADJ 7 cm4 cm

q

ADJ

OPPHYP

) x =5:6

cos 70±

) x ¼ 16:37

) x =27

tan 50±

) x ¼ 22:66

) x =12

tan 49

) x ¼ 10:43

) a =15

cos 27±

) a =45

sin 65±

) a ¼ 49:65

) b =45

tan 65±

) b ¼ 20:98

±

204 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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c cos µ =ADJ

HYP

=6

9=

2

3

) µ = cos¡1( 2

3)

) µ ¼ 48:2±

d sin µ =OPP

HYP

=3:2

5:6

) µ = sin¡1( 3:2

5:6)

) µ ¼ 34:8±

e tan µ =OPP

ADJ

=2:7

3:1

) µ = tan¡1( 2:7

3:1)

) µ ¼ 41:1±

f sin µ =OPP

HYP

=3

4

) µ = sin¡1( 3

4)

) µ ¼ 48:6±

g cos µ =ADJ

HYP

=4:7

5:2

) µ = cos¡1( 4:7

5:2)

) µ ¼ 25:3±

h tan µ =OPP

ADJ

=3:4

4:5

) µ = tan¡1( 3:4

4:5)

) µ ¼ 37:1±

i sin µ =OPP

HYP

=7:2

12:4

) µ = sin¡1( 7:2

12:4)

) µ ¼ 35:5±

9 cm6 cm

qHYP

OPP

ADJ

3 2. m

5 6. m

q

ADJOPP

HYP

2 7. km3 1. km

�

HYP

ADJ OPP

4 m3 m

�

HYP

OPP

ADJ

4 7. km

5 2. km

�

HYP

ADJ

OPP3 4. m 4 5. m

�

ADJOPP

HYP

7 2. mm

12 4. mm

�

HYP

ADJOPP

8 cm 5 cmf

q

x cmADJ

HYP

OPP

4 8. m

3 5. m

b

ax m

OPP

HYP

ADJ

9 45. km

12 06. km

a b

x km

HYP

OPPADJ

2 a sin µ =OPP

HYP=

5

8

) µ = sin¡1( 5

8)

) µ ¼ 38:7±

cos µ =ADJ

HYP=

x

8

) x = 8 cos µ

) x ¼ 8 cos 38:7±

) x ¼ 6:24

Á = 90± ¡ µ

¼ 90± ¡ 38:7± ¼ 51:3±

b tan® =OPP

ADJ=

4:8

3:5

) ® = tan¡1( 4:8

3:5)

) ® ¼ 53:902±

) ® ¼ 53:9±

sin® =OPP

HYP=

4:8

x

) x =4:8

sin®

) x ¼ 4:8

sin 53:902±

) x ¼ 5:94¯ = 90± ¡ ®

¼ 90± ¡ 53:9± ¼ 36:1±

c cos a =ADJ

HYP=

9:45

12:06

) a = cos¡1¡

9:45

12:06

¢) a ¼ 38:410±

) a ¼ 38:4±

sin a =OPP

HYP=

x

12:06

) x = 12:06 sina

) x ¼ 12:06 sin 38:410±

) x ¼ 7:49

b = 90± ¡ ®

¼ 90± ¡ 38:4± ¼ 51:6±

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 205

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3 a x2 + 52 = 82 fPythagorasg) x2 + 25 = 64

) x2 = 39

) x =p39 fas x > 0g

) x ¼ 6:24

b x2 = 3:52 + 4:82 fPythagorasg) x2 = 35:29

) x =p35:29 fas x > 0g

) x ¼ 5:94

c x2 + 9:452 = 12:062 fPythagorasg) x2 = 12:062 ¡ 9:452

) x2 = 56:1411

) x =p56:1411 fas x > 0g

) x ¼ 7:49

4 a sin µ =OPP

HYP

=8:1

7:9

) µ = sin¡1¡

8:1

7:9

¢This has no solution.

The triangle is impossible,

as the hypotenuse is shorter

than one of the other sides.

b cos µ =ADJ

HYP

=14

12=

7

6

) µ = cos¡1¡

7

6

¢This has no solution.

The triangle is impossible,

as the hypotenuse is shorter

than one of the other sides.

c cos µ =ADJ

HYP

=8:6

8:6= 1

) µ = cos¡1 (1)

) µ = 0±

The triangle does not really

exist, as one angle is 0±.

EXERCISE 10B.1

1 Let the height of the cliff be x m.

tan µ =OPP

ADJ

) tan 25± =x

235

) x = 235 tan 25±

) x ¼ 109:6

The cliff is about 110 m high.

2 Let the angle be µ.

cos µ =ADJ

HYP

=4:2

5

) µ = cos¡1¡

4:2

5

¢) µ ¼ 32:9±

The ladder makes an angle of 32:9± with the wall.

8 1. m

7 9. mq

HYP

OPP

ADJ

14 mm

12 mm

q

HYP

ADJOPP

8 6. km

8 6. km

q

HYP

ADJ

OPP

25°235 m

cliff

mx

ladderm5

wall

4 2. m

q

206 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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3

4

5

6

7

8

6°boat x m

25 m

lighthouse

28°

3 m

x m

6°

q

x m

80 m

boat

4°

4 km

x km

q

0.6 m5 m

x m

40°

2 5. m

B

A

Suppose the boat is x m from the lighthouse.

tan µ =OPP

ADJ

) tan 6± =25

x

) x =25

tan 6±

) x ¼ 237:9

The boat is about 238 m from the lighthouse.

Let the longer side be x m long.

cos µ =ADJ

HYP

) cos 28± =x

3

) x = 3cos 28±

) x ¼ 2:65

The longer side is about 2:65 m.

Suppose the boat is x m from the cliff.

µ = 6± falternate anglesgNow tan µ =

OPP

ADJ

) tan 6± =80

x

) x =80

tan 6±

) x ¼ 761:1

The boat is about 761 m out to sea.

Suppose the train gains x km in altitude.

tan µ =OPP

ADJ

) tan 4± =x

4

) x = 4 tan 4±

) x ¼ 0:2797

The train gains about 280 m in altitude.

sin µ =OPP

HYP

=0:6

5

) µ = sin¡1¡

0:6

5

¢) µ ¼ 6:89±

Suppose the strut is attached x m down the wall.

cos µ =ADJ

HYP

) cos 40± =x

2:5

) x = 2:5 cos 40±

) x ¼ 1:9151

The strut is attached about 1:92 m down the wall.

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 207

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9

The total height = 15 tan 25± +15

¼ 23:5 m

10

11 a Suppose P is x cm from the centre of the circle.

) x2 = 42 + 102 fPythagorasg) x2 = 116

) x =p116 fas x > 0g

) x ¼ 10:77 cm

P is about 10:8 cm from the centre of the circle.

b Suppose the angle is µ.

tan µ =OPP

ADJ=

4

10=

2

5

) µ = tan¡1¡

2

5

¢) µ ¼ 21:8±

The angle between the tangent and [OP] is about

21:8±.

12 The perpendicular from the centre of a circle to a chord,

bisects the chord. So, AM = BM = 4 cm.

In 4s OAM and OBM:

² OA = OB fequal radiig² AM = BM

² OM is common.

) triangles OAM and OBM are congruent fSSSg) AbOM = BbOM = µ

Now sin µ =OPP

HYP=

4

5

) µ = sin¡1¡

4

5

¢

x my m

25°

15 m

x m 20 m

55°

x cm

O

P10 cm

4 cm

q

A

B

MO

4 cm

q

q

5 cm

Suppose the goal post snapped x m above the ground,

and was (x+ y) m high in total.

tan µ =OPP

ADJ

) tan 25± =x

15

) x = 15 tan 25±

cos µ =ADJ

HYP

) cos 25± =15

y

) y =15

cos 25±

Suppose each cable is x m long.

sin µ =OPP

HYP

) sin 55± =20

x

) x =20

sin 55±

The total length of cable = 3£ 20

sin 55± ¼ 73:2 m.

) AbOB = 2 sin¡1¡

4

5

¢) AbOB ¼ 106±

cos 25±

208 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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13 The diagonals of a rhombus bisect each other at right

angles.

They also bisect the angles of the rhombus,

so µ = 1

2£ 76± = 38±.

Let a half of the longer diagonal be x cm.

Now cos µ =ADJ

HYP

) cos 38± =x

10

) x = 10 cos 38±

) the longer diagonal is 20 cos 38± ¼ 15:8 cm long.

14 The altitude of an isosceles triangle bisects the base and

bisects the angle at the apex.

) µ = 1

2£ 120± = 60±

Suppose the triangle is x m high.

Now tan µ =OPP

ADJ

) tan 60± =4:5

x

) the garage is about 3:5 + 2:60 ¼ 6:10 m high.

15 Suppose the tree is x m high.

tan µ =OPP

ADJ

) tan 70± =x

6

) x = 6 tan 70±

) x ¼ 16:5

The gardener needs to make 4 cuts to divide the tree up.

He also needs 1 cut to fell the tree, so that makes 5 cuts

in total.

16 The aeroplane is travelling at

200 km h¡1 =200£ 1000

60£ 60m s¡1

= 500

9m s¡1

) in the first 10 seconds it travels 5000

9m.

Suppose the aeroplane has altitude x m.

Now sin µ =OPP

HYP

) sin 27± =x

5000

9

) x = 5000

9sin 27±

) x ¼ 252:2

The plane has altitude 252 m.

q

q

x cm

10 cm

3 5. m

9 m

q q

4 5. m

x m

x m

70°70°

6 m

27°

x m

) x =4:5

tan 60± ¼ 2:60

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 209

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17 We assume the plane is travelling with

constant speed x m min¡1, and altitude

a m.

) every 2 minutes the plane travels

2x m.

Now tan µ =OPP

ADJ

) tan 27± =a

2x

) a = 2x tan 27±

18 The altitude of an isosceles triangle bisects the base.

Suppose the base is 6x m long.

) half the base is 3x m

and the other sides are 2

3£ 6x = 4x m.

Let the base angles be µ.

Now cos µ =ADJ

HYP=

3x

4x=

3

4

) µ = cos¡1¡

3

4

¢) µ ¼ 41:4±

The base angles are about 41:4±.

19 The altitude of an isosceles triangle bisects the base.

Suppose half the base is x cm, the altitude is h cm, and

the base angles are µ.

Now tan µ =OPP

ADJ=

h

x

) h = x tan µ

) the triangle has area = 1

2£ 2x£ h

= hx

= x2 tan µ cm2

In this case, both triangles have base 28 cm. ) 2x = 28

) x = 14

The first triangle has base angles 24±, and area 1

3that of the second triangle.

) 142 £ tan 24± = 1

3£ 142 £ tan µ

) tan µ = 3 tan 24±

) µ = tan¡1(3 tan 24±)

) µ ¼ 53:2±

The base angles are about 53:2±.

q q

3x m

4x m

q q

x cm

h cm

Suppose the angle of elevation after 4 minutes is Á.

) tanÁ =a

4x

So, tanÁ =2x tan 27±

4x

) tanÁ = 1

2tan 27±

) Á = tan¡1( 1

2tan 27±)

) Á ¼ 14:29±

The angle of elevation is about 14:3±.

f

a m

2x m

observer

27°

210 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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100 m

x m

22°

100 m

22° 40°

y m

b m

x m

B

C

D

56°

60°

30 mA

20 a Let the distance be x m.

tan µ =OPP

ADJ

) tan 22± =100

x

) x =100

tan 22±

) x ¼ 247:5

) the distance is about 248 m.

b Suppose the point is y m from the building.

tan µ =OPP

ADJ

) tan 40± =100

y

) y =100

tan 40±

) the point has moved

100

tan 22± ¡ 100

tan 40± ¼ 128 m closer to the building.

21 Suppose the building is b m high and the flagpole is x m high.

tan µ =OPP

ADJ

) tan 56± =b

30and tan 60± =

x+ b

30

) tan 60± =x

30+ tan 56±

)x

30= tan 60± ¡ tan 56±

) x = 30(tan 60± ¡ tan 56±)

) x ¼ 7:48

The flagpole is about 7:48 m high.

22

22°19°

pole B pole A

100 m

AB

h m

M

Suppose the river is x m wide, so AM = x m.

Suppose also that the poles are both h m high.

Now tan µ =OPP

ADJ

) tan 19± =h

BMand tan 22± =

h

x

) h = BM £ tan 19± and h = x tan 22±

So, BM =x tan 22±

tan 19±

Now BM2 = AM2 + AB2 fPythagorasg

)

³x tan 22±

tan 19

´2

= x2 + 1002

) x2

µ³tan 22±

tan 19±

´2

¡ 1

¶= 1002

) x2 =10000¡

tan 22±tan 19±

¢2 ¡ 1

) x2 ¼ 26 538:3

) x ¼ 162:9 The river is about 163 m wide.

±

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 211

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23

24

EXERCISE 10B.2

1 a b c d

23° 37°

750 m x m

h m

y

x3 6

2

4 B

x

q1

q2A

D

E

C

N

O136°

N

O

240°

N

O

51°

N

O 327°

Let AC = x units

µ1 = µ2 fcorresponding anglesg

In 4ADE, tan µ =OPP

ADJ=

DE

AD=

3

2

) µ = tan¡1( 3

2) .... (1)

In 4ABC, sin µ =OPP

HYP=

AC

AB=

x

2

) x = 2 sin µ

) x = 2 sin¡tan¡1( 3

2)¢

fusing (1)g) x ¼ 1:66

The lines are about 1:66 units apart.

2 a Since the bearing of P from Q is 054±,

the bearing of Q from P is 054± + 180± = 234±.

b Since the bearing of P from Q is 113±,

the bearing of Q from P is 113± + 180± = 293±.

c Since the bearing of P from Q is 263±,

the bearing of Q from P is 263± ¡ 180± = 083±.

d Since the bearing of P from Q is 304±,

the bearing of Q from P is 304± ¡ 180± = 124±.

Let the volcano be h m high.

Suppose the second measurement point is x m

horizontally from its rim.

tan µ =OPP

ADJ

) tan 23± =h

x+ 750and tan 37± =

h

x

) x+ 750 =h

tan 23± and x =h

tan 37±

) x =h

tan 23± ¡ 750 and x =h

tan 37±

So,h

tan 23± ¡ 750 =h

tan 37±

) h

³1

tan 23± ¡ 1

tan 37±

´= 750

) h =750

1

tan 23± ¡ 1

tan 37±

) h ¼ 728:999

The volcano is about 729 m high.

212 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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N

q

14 km

9 km

N

A

q

d km

6 km

5 km

start B

N

20 km

qd km

153°

N

d km

40°60 km

3 a i The bearing of B from A is 041±.

ii The bearing of C from B is 142±.

iii The bearing of B from C is

142± + 180± = 322±.

iv The bearing of A from C is 279±

) the bearing of C from A is

279± ¡ 180± = 099±.

v The bearing of A from B is

41± + 180± = 221±.

vi The bearing of A from C is 279±.

b i The bearing of B from A is 027±.

ii The bearing of C from B is 151±.

iii The bearing of B from C is

151± + 180± = 331±.

iv The bearing of A from C is 246±

) the bearing of C from A is

246± ¡ 180± = 066±.

v The bearing of A from B is

27± + 180± = 207±.

vi The bearing of A from C is 246±.

4 tan µ =OPP

ADJ=

9

14

) µ = tan¡1( 9

14)

) µ ¼ 32:7±

The bearing of the finishing position from the starting

point is 90± + 32:7± ¼ 123±.

5 After 30 mins, runner A has travelled 1

2£ 10 = 5 km north,

and runner B has travelled 1

2£ 12 = 6 km east.

d2 = 52 + 62 fPythagorasg) d2 = 25 + 36

) d2 = 61

) d ¼ 7:81 fas d > 0g

tan µ =OPP

ADJ=

6

5

) µ = tan¡1( 6

5)

) µ ¼ 50:2±

) 180± ¡ µ ¼ 129:8±

So, runner B is now 7:81 km from runner A, on the bearing 130±.

6 Let the distance she walked be d km.

µ = 180± ¡ 153± = 27±

cos µ =ADJ

HYP

) cos 27± =20

d

) d =20

cos 27±

) d ¼ 22:4

The hiker walked about 22:4 km.

7 Suppose the boat is d km east of its starting point.

sin µ =OPP

HYP

) sin 40± =d

60

) d = 60 sin 40±

) d ¼ 38:6

The boat is about 38:6 km east of its starting point.

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 213

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8 Suppose the aeroplane travelled d km.

µ = 360± ¡ 295± = 65±

sin µ =OPP

HYP

) sin 65± =200

d

) d ¼ 220:7

The aeroplane travelled about 221 km.

9 a Suppose the trawler T is d km from P.

114± ¡ 24± = 90±, so the trawler turned a right

angle.

So, d2 = 202 + 302 fPythagorasg) d2 = 400 + 900

) d2 = 1300

) d ¼ 36:06

tan µ =OPP

ADJ=

20

30=

2

3

) µ = tan¡1( 2

3)

) µ ¼ 33:7±

) µ + 24± ¼ 57:7±

The trawler is about 36:1 km from P, on the bearing

057:7±.

b 057:7± + 180± = 237:7±

) the trawler must sail on the bearing 238±.

EXERCISE 10C.1

1 a

x2 = 152 + 152 fPythagorasg) x2 = 450

) x =p450 = 15

p2

) x ¼ 21:21

So, EG ¼ 21:2 cm

b

tan µ =OPP

ADJ=

15

15p2=

1p2

) µ = tan¡1( 1p2)

) µ ¼ 35:3±

So, AbGE ¼ 35:3±

2 a x2 = 52 + 82 fPythagorasg) x2 = 25 + 64 = 89

) x =p89

) x ¼ 9:43

HX ¼ 9:43 cm

200 km

d km

295°

q

N

N

T

P

N30 km

24° d km

20 km

114°

q

x cm15 cm

G

F

H

E

GH

F

BA

DC

Eq

E F

G

CD

A

H

B

X

Y

6 cm

4 cm

5 cm 5 cm

8 cmx cm

E X

H

E G

A

15 cm

15~`2 cm

q

) d =200

sin 65±

214 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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b

tan µ =OPP

ADJ=

6p89

) µ = tan¡1( 6p89

)

) µ ¼ 32:5±

So, DbXH ¼ 32:5±

c

d2 = 42 + 102

fPythagorasg) d2 = 16 + 100 = 116

) d =p116

) d ¼ 10:8

HY ¼ 10:8 cm

d

tan µ =OPP

ADJ=

6p116

) µ = tan¡1( 6p116

)

) µ ¼ 29:1±

So, DbYH ¼ 29:1±

3

a

x2 = 42 + 82 fPythagorasg) x2 = 16 + 64 = 80

) x =p80

) x ¼ 8:94

DF ¼ 8:94 cm

b

tan µ =OPP

ADJ=

3p80

) µ = tan¡1( 3p80

)

) µ ¼ 18:5±

So, AbFD ¼ 18:5±

4

~`8`9 cm

6 cm

qH X

D

4 cmd cm

10 cmH G

Y H Y

D

6 cm

~`1`1`6 cm

q

A

B

C

F

E

D

8 cm

4 cm

3 cm

E F

D

4 cmx cm

8 cm D F

A

3 cm

~`8`0 cm

q

x cm

A

B

40°

y cm

25 cm

z cm

C

B

35°

25 40°_tan_

Now cos 40± =ADJ

HYP=

25

x

) x =25

cos 40±

Also, tan 40± =OPP

ADJ=

y

25

) y = 25 tan 40±

sin 35± =OPP

HYP=

25 tan 40±

z

) z =25 tan 40±

sin 35±

So, the total length of wood = AB + BC

=25

cos 40± +25 tan 40±

sin 35±

¼ 69:2 cm

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 215

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5 a

EXERCISE 10C.2

1 a i [GF] ii [HG] iii [HF] iv [GM]

b i [MA] ii [MN]

c i [CD] ii [DE] iii [DF] iv [DX]

2 a i DbEH ii CbEG iii AbGE iv BbXF

b i PbYS ii QbWR iii QbXR iv QbYR

c i AbQX ii AbYX

3 a i

The required angle is DbEH.

4DEH is a right angled isosceles

triangle.

) the angle is 45±.

ii

The required angle is b(FH)2 = 102 + 102 fPythagorasg) FH =

p200 = 10

p2 cm

tan µ =OPP

ADJ=

10

10p2=

1p2

) µ = tan¡1( 1p2) ¼ 35:3±

) the angle is about 35:3±.

A

B

C

M 12 mq

12 m

6~`2 m

qA M

C

x m

qA M

C

xp2

m

A B

F

GH

D

E

C

10 m

AB2 = 122 + 122 fPythagorasg) AB2 = 144£ 2

) AB = 12p2 m fas AB > 0g

) AM = 6p2 m

cos µ =ADJ

HYP=

6p2

12

) cos µ =

p2

2=

1p2

) µ = cos¡1( 1p2)

) µ = 45±

The angle between the slant edge and

a base diagonal is 45±.

b Suppose the sides of the pyramid are x m long.

Then AB2 = x2 + x2 fPythagorasg= 2x2

) AB = xp2 m fAB > 0g

) AM =xp2

2m

=xp2

m

cos µ =ADJ

HYP=

xp2

x=

1p2

) µ = cos¡1( 1p2)

) µ = 45±

So, the angle is always 45±.

DFH.

A B

F

GH

D

E

C

10 m

q

216 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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iii iv

The required angle is AbXE.

(EX)2 = 102 + 52 fPythagorasg) (EX)2 = 125

) EX =p125 = 5

p5 cm

tan¯ =OPP

ADJ=

10

5p5=

2p5

) ¯ = tan¡1( 2p5) ¼ 41:8±

) the angle is about 41:8±.

b i

The required angle is PbUT.

tan µ =OPP

ADJ=

4

10=

2

5

) µ = tan¡1( 2

5) ¼ 21:8±

) the angle is about 21:8±.

ii

The required angle is PbVT.

(TV)2 = 102 + 62 fPythagorasg) TV =

p136 cm

tan® =OPP

ADJ=

4p136

) ® = tan¡1( 4p136

) ¼ 18:9±

) the angle is about 18:9±.

iii

The required angle is SbXW.

(WX)2 = 102 + 32 fPythagorasg) WX =

p109 cm

tan¯ =OPP

ADJ=

4p109

) ¯ = tan¡1( 4p109

) ¼ 21:0±

) the angle is about 21:0±.

10 cm

D

H Xa

5 cm

W V

U

QP

T

RS

6 cm

4 cm

10 cm

q

10 cm

G

F

B

C

A

D

H

E

X5 cm

b

W V

U

QP

TR

S

6 cm

4 cm

10 cm

a

W V

U

QP

T

RS

6 cm

4 cm

10 cm

b 3 cmX

The required angle is DbXH.

tan® =OPP

ADJ=

10

5= 2

) ® = tan¡1(2) ¼ 63:4±

) the angle is about 63:4±.

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 217

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c i

The required angle is KbOL.

tan µ =OPP

ADJ=

3

4

) µ = tan¡1( 3

4) ¼ 36:9±

) the angle is about 36:9±.

ii

The required angle is JbXM.

(MX)2 = 22 + 42 fPythagorasg) MX =

p20 cm

tan® =OPP

ADJ=

3p20

) ® = tan¡1( 3p20

) ¼ 33:9±

) the angle is about 33:9±.

iii

d i

The required angle is XbDM.

Now (BD)2 = 102 + 122 fPythagorasg) BD =

p244 cm

) MD = 1

2

p244 cm

cos µ =ADJ

HYP=

MD

XD=

1

2

p244

15

) µ = cos¡1(p

244

30) ¼ 58:6±

) the angle is about 58:6±.

ii

The required angle is MbYX.

Now MY = 6 cm

and (XY)2 + 52 = 152 fPythagorasg) (XY)2 = 200

) XY = 10p2 cm

cos® =ADJ

HYP=

MY

XY=

6

10p2

) ® = cos¡1( 3

5p

2) ¼ 64:9±

) the angle is about 64:9±.

EXERCISE 10D

1 a 1 b 0 c 0 d 1 e 0 f ¡1

g ¡1 h 0 i 1 j 0 k 0 l 1

qO L

K

4 m

3 m

a

ON

J K

LM

2 m

4 m

3 m

X

ON

J K

LM

4 m

4 m

3 m

Y

2 m b

D

B

X

M 10 cm

15 cm

12 cm

q

C

A

D

B

X

M5 cm

15 cm

12 cmC

A

a

Y

The required angle is KbYL.

(YL)2 = 22 + 42 fPythagorasg) YL =

p20 cm

tan¯ =OPP

ADJ=

3p20

) ¯ = tan¡1( 3p20

) ¼ 33:9±

) the angle is about 33:9±.

218 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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2 a 0:64 b 0:77 c ¡0:34 d 0:94 e 0:17 f ¡0:98

g ¡0:77 h ¡0:64 i 0:77 j ¡0:64 k 0:87 l ¡0:5

4

5

6 a

cos µ = 1

2

) µ = cos¡1( 1

2)

) µ = 60±

b

cos µ = ¡ 1

2

) µ = cos¡1(¡1

2)

) µ = 120±

c

sin µ = 1

2

) µ = sin¡1( 1

2)

or 180± ¡ sin¡1( 1

2)

) µ = 30± or 150±

7 a

cos µ = 0:672

) µ = cos¡1(0:672)

) µ ¼ 47:8±

b

cos µ = ¡0:672

) µ = cos¡1(¡0:672)

) µ ¼ 132:2±

c

cos µ = 0:138

) µ = cos¡1(0:138)

) µ ¼ 82:1±

y

x

( , )cos__ sin__q q

180° + q

q

0.138

x

y

x

y

Qw x

y

-Qw x

y

Qw

x

y

0.672

x

y

-0.672

y

x

( , )cos__ sin__q q

q

-q

180° - q

a The y-coordinates for µ and 180±¡ µ are the same.

) sin(180± ¡ µ) = sin µ

b The x-coordinates for µ and 180± ¡ µ are the

negatives of each other.

) cos(180± ¡ µ) = ¡ cos µ

c The x-coordinates for µ and ¡µ are the same.

) cos(¡µ) = cos µ

d The y-coordinates for µ and ¡µ are the negatives

of each other.

) sin(¡µ) = ¡ sin µ

a The y-coordinates for µ and 180± + µ are the

negatives of each other.

) sin(180± + µ) = ¡ sin µ

b The x-coordinates for µ and 180± + µ are the

negatives of each other.

) cos(180± + µ) = ¡ cos µ

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 219

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d

cos µ = ¡0:138

) µ = cos¡1(¡0:138)

) µ ¼ 97:9±

e

sin µ = 0:317

) µ = sin¡1(0:317)

or 180± ¡ sin¡1(0:317)

) µ ¼ 18:5± or 161:5±

f

sin µ = 0:887

) µ = sin¡1(0:887)

or 180± ¡ sin¡1(0:887)

) µ ¼ 62:5± or 117:5±

g

cos µ = 0:077

) µ = cos¡1(0:077)

) µ ¼ 85:6±

h

cos µ = ¡0:369

) µ = cos¡1(¡0:369)

) µ ¼ 111:7±

i

sin µ = 0:929

) µ = sin¡1(0:929)

or 180± ¡ sin¡1(0:929)

) µ ¼ 68:3± or 111:7±

8 a

tan 0± = 0

b

tan 45± = 1

c

As µ approaches 90±, the

length of the tangent gets

larger and larger.

) tan µ approaches

infinity.

9 a i OQ = cos µ

b

x

y

-0.138

x

y

0 317.

x

y0.887

x

y

0.077

x

y

-0.369

x

y0.929

x

y

1P x

y

45°

1

T

1

P

x

y

q1

P

O A

T

Q

P

cos__q1

sin__q

tan__q

ii PQ = sin µ iii AT = tan µ

In 4s OPQ and OTA:

² PbOQ is common

² ObQP = ObAT = 90±

) the triangles are similar.

)AT

QP=

AO

QO

)tan µ

sin µ=

1

cos µ

) tan µ =sin µ

cos µ

220 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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EXERCISE 10E

1 a Area = 1

2ab sinC

= 1

2£ 12£ 13£ sin 45±

¼ 55:2 cm2

b Area = 1

2ab sinC

= 1

2£ 28£ 25£ sin 82±

¼ 347 km2

c Area = 1

2ab sinC

= 1

2£ 6:4£ 7:8£ sin 112±

¼ 23:1 cm2

d Area = 1

2ab sinC

= 1

2£ 32£ 27£ sin 84±

¼ 430 m2

e Area = 1

2ab sinC

= 1

2£ 10:6£ 12:2£ sin 125±

¼ 53:0 cm2

f Area = 1

2ab sinC

= 1

2£ 1:43£ 1:65£ sin 78±

¼ 1:15 m2

2 The triangles are congruent. fSSSg) the total area = 2£ 1

2ab sinC

= 2£ 1

2£ 6:4£ 8:7£ sin 64±

¼ 50:0 cm2

3 Area = 1

2ac sinB

) 150 = 1

2£ x£ 14£ sin 75±

) x =300

14 sin 75±

) x ¼ 22:2

4 Area = 1

2ab sinC

) 30 = 1

2£ 10£ 12£ sin µ where 0± 6 µ 6 180±

) sin µ = 60

120= 1

2

) µ = sin¡1( 1

2) or 180± ¡ sin¡1( 1

2)

) µ = 30± or 150±

64°

6 4. cm

8.7 cm

10 m

12 m

qQ R

P

x

y

Qw

10 Using question 4, tan(180± ¡ µ) =sin(180± ¡ µ)

cos(180± ¡ µ)

=sin µ

¡ cos µ

= ¡ sin µ

cos µ

= ¡ tan µ

Check: tan 23± ¼ 0:4245 tan(180± ¡ 23±) = tan 157± ¼ ¡0:4245 X

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 221

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5

6 a i Area = 1

2bc sinA

ii Area = 1

2ab sinC

b Equating the areas in a, 1

2ab sinC = 1

2bc sinA

) a sinC = c sinA

)a

c=

sinA

sinC

EXERCISE 10F

A

C

B

13 cm 17 cm

q

Area = 1

2ac sinB

) 73:4 = 1

2£ 13£ 17£ sin µ where 0± 6 µ 6 180±

) sin µ ¼ 0:6643

) µ ¼ sin¡1(0:6643) or 180± ¡ sin¡1(0:6643)

) AbBC ¼ 41:6± or 138:4±

1 a Using the sine rule,

x

sin 32± =15

sin 46±

) x =15 sin 32±

sin 46±

) x ¼ 11:1

b Using the sine rule,

x

sin 108± =9

sin 48±

) x =9 sin 108±

sin 48±

) x ¼ 11:5

c Using the sine rule,

x

sin 55± =6:3

sin 84±

) x =6:3 sin 55±

sin 84±

) x ¼ 5:19

2 a Using the sine rule,

a

sinA=

b

sinB

) a =b sinA

sinB

) a =18 sin 65±

sin 35±

) a ¼ 28:4 cm

b Using the sine rule,

b

sinB=

c

sinC

Now B = 180± ¡A¡ C

= 180± ¡ 72± ¡ 27±

= 81±

So,b

sin 81± =24

sin 27±

) b =24 sin 81±

sin 27±

) b ¼ 52:2 cm

c Using the sine rule,c

sinC=

a

sinA

Now A = 180± ¡B ¡ C

= 180± ¡ 25± ¡ 42±

= 113±

So,c

sin 42± =7:2

sin 113±

) c =7:2 sin 42±

sin 113±

) c ¼ 5:23 cm

3 a Using the sine rule,

sin µ

14:8=

sin 38±

17:5

) sin µ =14:8 sin 38±

17:5

Now sin¡1

³14:8 sin 38±

17:5

´¼ 31:4±

) since µ could be acute or obtuse,

µ ¼ 31:4± or (180¡ 31:4)± ¼ 148:6±

We reject µ ¼ 148:6± as

148:6± + 38± > 180±

) µ ¼ 31:4±

b Using the sine rule,

sin µ

35=

sin 54±

29

) sin µ =35 sin 54±

29

Now sin¡1

³35 sin 54±

29

´¼ 77:5±

) since µ could be acute or obtuse,

µ ¼ 77:5± or (180¡ 77:5)± ¼ 102:5±

There is insufficient information to determine

the exact shape of the triangle.

222 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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15.1 cm

12.6 cm

65°

AA C

B

38.4 cm

27.6 cm

43°A

B

C

B

5.5 km

71°A

C

B

4 1. km

C

c Using the sine rule,

sin µ

6:4=

sin 15±

2:4

) sin µ =6:4 sin 15±

2:4

Now sin¡1

³6:4 sin 15±

2:4

´¼ 43:6±

) since µ could be acute or obtuse, µ ¼ 43:6± or (180¡ 43:6)± ¼ 136:4±

There is insufficient information to determine the exact shape of the triangle.

4 a Using the sine rule,

sinA

a=

sinB

b

)sinA

12:6=

sin 65±

15:1

) sinA =12:6 sin 65±

15:1

Now sin¡1

³12:6 sin 65±

15:1

´¼ 49:1±

) since A could be acute or obtuse, A ¼ 49:1± or (180¡ 49:1)± ¼ 130:9±

We reject A ¼ 130:9±, as 130:9± + 65± > 180±

) A ¼ 49:1±

b Using the sine rule,

sinB

b=

sinC

c

)sinB

38:4=

sin 43±

27:6

) sinB =38:4 sin 43±

27:6

Now sin¡1

³38:4 sin 43±

27:6

´¼ 71:6±

) since B could be acute or obtuse, B ¼ 71:6± or (180¡ 71:6)± ¼ 108:4±

There is insufficient information to determine the exact shape of the triangle.

c Using the sine rule,

sinC

c=

sinA

a

)sinC

4:1=

sin 71±

5:5

) sinC =4:1 sin 71±

5:5

Now sin¡1

³4:1 sin 71±

5:5

´¼ 44:8±

) since C could be acute or obtuse, C ¼ 44:8± or (180¡ 44:8)± ¼ 135:2±

We reject C ¼ 135:2±, as 135:2± + 71± > 180±

) C ¼ 44:8±

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 223

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EXERCISE 10G

1 a

b

c

2 Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) 2bc cosA = b2 + c2 ¡ a2

) cosA =b2 + c2 ¡ a2

2bc

) A = cos¡1

µ92 + 142 ¡ 112

2£ 9£ 14

¶) A ¼ 51:75±

Also using the cosine rule,

b2 = a2 + c2 ¡ 2ac cosB

) 2ac cosB = a2 + c2 ¡ b2

) cosB =a2 + c2 ¡ b2

2ac

) B = cos¡1

µ112 + 142 ¡ 92

2£ 11£ 14

¶) B ¼ 39:98±

) C = 180± ¡A¡B

¼ 180± ¡ 51:75± ¡ 39:98±

¼ 88:27±

So, BbAC ¼ 51:8±, AbBC ¼ 40:0±,

and AbCB ¼ 88:3±.

3 a The smallest angle is opposite the shortest side.

) the smallest angle is A.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) cosA =b2 + c2 ¡ a2

2bc

) A = cos¡1

µ112 + 132 ¡ 92

2£ 11£ 13

¶) A ¼ 43:0±

The smallest angle is about 43:0±.

A

B

C

21 cm

15 cm

a cm

98°

r km

4.8 kmQ

P

R38°

6 7. km

K

L M

10 3. ml m

14 8. m

72°

A B

C

A B

C

9 m 11 m

14 m

C A

B

9 cm 11 cm

13 cm

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) a =p

212 + 152 ¡ 2£ 21£ 15£ cos 98±

) a ¼ 27:5

) [BC] is about 27:5 cm long.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) r =p

4:82 + 6:72 ¡ 2£ 4:8£ 6:7£ cos 38±

) r ¼ 4:15

) [PQ] is about 4:15 km long.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) l =p

10:32 + 14:82 ¡ 2£ 10:3£ 14:8£ cos 72±

) l ¼ 15:2

) [KM] is about 15:2 m long.

224 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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b The largest angle is opposite the longest side.

) the largest angle is A.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) cosA =b2 + c2 ¡ a2

2bc

) A = cos¡1

µ32 + 52 ¡ 72

2£ 3£ 5

¶) A = 120±

The largest angle is 120±.

4 a Using the cosine rule,

a2 = c2 +m2 ¡ 2cm cos µ

) cos µ =c2 +m2 ¡ a2

2cm

b Using the cosine rule,

b2 = c2 +m2 ¡ 2cm cos(180± ¡ µ)

) cos(180± ¡ µ) =c2 +m2 ¡ b2

2cm

c Since cos(180± ¡ µ) = ¡ cos µ,

c2 +m2 ¡ b2

2cm= ¡c2 +m2 ¡ a2

2cm

) c2 +m2 ¡ b2 = ¡c2 ¡m2 + a2

) a2 + b2 = 2m2 + 2c2 as required

d i Using Apollonius’ median theorem,

122 + 92 = 2x2 + 2£ 52

) 2x2 = 175

) x2 = 175

2

) x ¼ 9:35 fas x > 0g

ii Using Apollonius’ median theorem,

82 + 102 = 2£ 82 + 2x2

) 2x2 = 36

) x2 = 18

) x ¼ 4:24 fas x > 0g

B C

A

3 cm 5 cm

7 cm

60°

A

B

C

10 cm

9 cm

x cm

5 a Using the cosine rule,

b2 = a2 + c2 ¡ 2ac cosB

) 92 = x2 + 102 ¡ 2x£ 10 cos 60±

) 81 = x2 + 100¡ 20x£ 1

2

) x2 ¡ 10x+ 19 = 0

b Using the quadratic formula,

x =¡(¡10)§

p(¡10)2 ¡ 4£ 1£ 19

2£ 1

) x =10§p

24

2) x = 5§

p6

) x ¼ 7:45 or 2:55

c Scale: 50%

2 55. cm

60°

A

C2

B

10 cm

9 cm

C1

7 45. cm

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 225

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EXERCISE 10H

1 Let AC be x km.

Using the sine rule,b

sinB=

c

sinC

Now AbCB = 180± ¡ 83± ¡ 59±

= 38±

2

3 Suppose AC is x km

so BC is 2x km.

Using the sine rule,sinB

b=

sinA

a

)sinB

x=

sin 50±

2x

) sinB = 1

2sin 50±

Now sin¡1( 1

2sin 50±) ¼ 22:5±

) B ¼ 22:5± or (180¡ 22:5)± ¼ 157:5±

We reject B ¼ 157:5± as 157:5± + 50± > 180±

) B ¼ 22:5±

) BbCA ¼ 180± ¡ 50± ¡ 22:5±

) BbCA ¼ 107:5±

4 a Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) cosA =b2 + c2 ¡ a2

2bc

) A = cos¡1

µ4072 + 3142 ¡ 2382

2£ 407£ 314

¶) A ¼ 35:686±

So, the angle at A is about 35:69±.

b Area = 1

2bc sinA

¼ 1

2£ 407£ 314£ sin 35:686±

¼ 37 275 m2

¼ 4 ha

Hazel’s property is about 4 hectares.

83° 59°

x km

C

A B10 3. km

524 m

23°

b m

AB

C

786 m

A

C

B

B

50°

2x km

x km

C is about 14:3 km from A.

AbBC = 180± ¡ 23± = 157± fangles on a linegUsing the cosine rule,

b2 = a2 + c2 ¡ 2ac cosB

) b =p

7862 + 5242 ¡ 2£ 786£ 524£ cos 157±

) b ¼ 1284:8

It is about 1285 m from A direct to C.

So,x

sin 59± =10:3

sin 38±

) x =10:3 sin 59±

sin 38±

) x ¼ 14:3

226 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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NN

N

127°

startA q

6 km 53°

x km

4 km

finish

f

B

C

5 Suppose the orienteer is x m from her starting point.

µ = 180± ¡ 32± = 148±

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) x =p

4502 + 6002 ¡ 2£ 450£ 600£ cos 148±

) x ¼ 1010

The orienteer is about 1010 m from her starting point.

6 Suppose the yacht is x km from its starting position,

and let µ and Á be the angles marked.

Á+ 127± = 180± fallied anglesg) Á = 53±

Using the cosine rule,

b2 = a2 + c2 ¡ 2ac cosB

) x =p

42 + 62 ¡ 2£ 4£ 6£ cos 106±

) x ¼ 8:0765

Also using the cosine rule, a2 = b2 + c2 ¡ 2bc cosA

) cosA =b2 + c2 ¡ a2

2bc

) µ ¼ cos¡1

µ8:07652 + 62 ¡ 42

2£ 8:0765£ 6

¶) µ ¼ 28:4±

) 127± ¡ µ ¼ 98:6±

So, the yacht is about 8:08 km on the bearing 098:6± from its starting point.

7 XbYZ = 146± ¡ 72± = 74±

Using the sine rule,

sinXbZY

XY=

sinXbYZ

XZ

) sinXbZY =9 sin 74±

14

) XbZY = sin¡1( 9

14sin 74±) or 180± ¡ sin¡1( 9

14sin 74±)

) XbZY ¼ 38:17± or 141:83±

But we reject XbZY ¼ 141:83± as 74± + 141:83± > 180±

So, XbZY ¼ 38:17±

µ + 72± = 180± fallied anglesg) µ = 108±

Now Á+ XbZY + µ = 360±

) Á ¼ 360± ¡ 108± ¡ 38:17±

¼ 213:83±

) X is on the bearing 214± from Z.

x m

32°

450 m

600 m

q

start

finish

N

N

N

146°

72°

Y

q

f

X

14 km

Z

9 km

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 227

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8 a

µ + 37± = 41± fexternal angle of triangleg) µ = 4±

So, AbTB = 4±

c d Let the mountain be z m high.

Now sin 41± =z

y

) z ¼ 10 353 sin 41±

) z ¼ 6792

So, the mountain is about 6790 m high.

e

tanÁ =h

x, so x =

h

tanÁ

Also, tan µ =h

x+ d

) x+ d =h

tan µ

) d =h

tan µ¡ x

) d =h

tan µ¡ h

tanÁ

) d = h

µ1

tan µ¡ 1

tanÁ

¶

f Using e, h =d

1

tan µ¡ 1

tanÁ

) h =1200

1

tan 37± ¡ 1

tan 41±

) h ¼ 6790 m as before.

9 µ1 = 238± ¡ 180± = 58±

µ2 = µ1 falternate anglesg) µ2 = 58±

) PbQR = 107± ¡ 58± = 49±

Let PR be x km.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) x =p

11:32 + 18:92 ¡ 2£ 11:3£ 18:9£ cos 49±

) x ¼ 14:3

37° 41°A C

T

x m

y m

1200 m

z m

B

q

q f

d x

h

N

N

238°

q111 3. km

107°

q2

Q

R

P

18 9. km

x km

b Let AT be x m.

Using the sine rule,

x

sinAbBT=

1200

sinAbTB

Now AbBT = 180± ¡ 41± = 139±

) x =1200 sin 139±

sin 4±

) x ¼ 11 286

So, AT is about 11:3 km.

Let BT be y m.

Using the sine rule,y

sinTbAB=

1200

sinAbTB

) y =1200 sin 37±

sin 4±

) y ¼ 10 353

So, BT is about 10:4 km.

So, R is about 14:3 km from the starting point P.

228 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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10 Let [AB] be x m long

) [DC] is (x+ 5) m long.

Using the cosine rule on each of the two triangles,

BD2 = x2 + 162 ¡ 2x£ 16 cos 120±

and BD2 = (x+ 5)2 + 252 ¡ 2(x+ 5)£ 25 cos 60±

Equating,

x2 + 256¡ 32x cos 120± = (x+ 5)2 + 625¡ 50(x+ 5) cos 60±

) x2 + 256 + 16x = x2 + 10x+ 650¡ 25(x+ 5)

) 256 + 16x = ¡15x+ 525

) 31x = 269

) x = 269

31

The fence has length 16+x+25+(x+5) ~ 63:4 m.

EXERCISE 10I.1

1 cos2 µ + sin2 µ = 1

) cos2 µ + ( 2

3)2 = 1

) cos2 µ + 4

9= 1

) cos2 µ = 5

9

) cos µ = §p

5

3

a If 0± < µ < 90±

then cos µ > 0

) cos µ =p

5

3

b If 90± < µ < 180±

then cos µ < 0

) cos µ = ¡p

5

3

2 cos2 µ + sin2 µ = 1

) cos2 µ + ( 1p3)2 = 1

) cos2 µ + 1

3= 1

) cos2 µ = 2

3

Now 90± < µ < 180±, so cos µ < 0

) cos µ = ¡p

2p3

) cos µ = ¡p

6

3

3 cos2 µ + sin2 µ = 1

) (¡ 3

5)2 + sin2 µ = 1

) 9

25+ sin2 µ = 1

) sin2 µ = 16

25

Now 90± < µ < 180±, so sin µ > 0

) sin µ = 4

5

EXERCISE 10I.2

1 a cos µ + cos µ = 2cos µ b 2 sin µ + 3 sin µ = 5 sin µ c 4 sin µ ¡ sin µ = 3 sin µ

d 5 sin µ ¡ 3 sin µ = 2 sin µ e 2 cos µ ¡ 5 cos µ = ¡3 cos µ f 12 cos µ ¡ 7 cos µ = 5 cos µ

2 a 5 sin2 µ + 5cos2 µ

= 5(sin2 µ + cos2 µ)

= 5£ 1

= 5

b ¡3 sin2 µ ¡ 3 cos2 µ

= ¡3(sin2 µ + cos2 µ)

= ¡3£ 1

= ¡3

c ¡ sin2 µ ¡ cos2 µ

= ¡1(sin2 µ + cos2 µ)

= ¡1£ 1

= ¡1

60°

120°

16 m

x m

25 m

D

B

C

A

(x + 5) m

y

x

We

y

x

-Et

y

x

1p3

d 7¡ 7 sin2 µ

= 7(1¡ sin2 µ)

= 7 cos2 µ

fcos2 µ + sin2 µ = 1g

e 6¡ 6 cos2 µ

= 6(1¡ cos2 µ)

= 6 sin2 µ

fcos2 µ + sin2 µ = 1g

f sin µ cos2 µ + sin3 µ

= sin µ(cos2 µ + sin2 µ)

= sin µ £ 1

= sin µ

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 229

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3 a (2 + sin µ)2

= 4 + 4 sin µ + sin2 µ

b (sin®¡ 3)2

= sin2 ®¡ 6 sin®+ 9

c (cos®¡ 4)2

= cos2 ®¡ 8 cos®+ 16

d (sin¯ + cos¯)2

= sin2 ¯ + 2 sin¯ cos¯ + cos2 ¯

= cos2 ¯ + sin2 ¯ + 2 sin¯ cos¯

= 1 + 2 sin¯ cos¯

e (sinÁ¡ cosÁ)2

= sin2 Á¡ 2 sinÁ cosÁ+ cos2 Á

= cos2 Á+ sin2 Á¡ 2 sinÁ cosÁ

= 1¡ 2 sinÁ cosÁ

f ¡(1¡ cos®)2

= ¡(1¡ 2 cos®+ cos2 ®)

= ¡1 + 2 cos®¡ cos2 ®

4 a 1¡ sin2 Á

= (1 + sinÁ)(1¡ sinÁ)

fa2 ¡ b2 = (a+ b)(a¡ b)g

b sin2 µ ¡ cos2 µ

= (sin µ + cos µ)(sin µ ¡ cos µ)

fa2 ¡ b2 = (a+ b)(a¡ b)gc cos2 ¯ ¡ 1

= (cos¯ + 1)(cos¯ ¡ 1)

fa2 ¡ b2 = (a+ b)(a¡ b)g

d 3 sin2 ¯ ¡ sin¯

= sin¯(3 sin¯ ¡ 1)

e 6 cosÁ+ 3 cos2 Á

= 3cosÁ(2 + cosÁ)

f 4 sin2 µ ¡ 2 sin µ

= 2 sin µ(2 sin µ ¡ 1)

g sin2 µ + 6 sin µ + 8

= (sin µ + 4)(sin µ + 2)

fx2 + 6x+ 8 = (x+ 4)(x+ 2)g

h 2 cos2 µ + 7cos µ + 6

= (2 cos µ + 3)(cos µ + 2)

f2x2 + 7x+ 6 = (2x+ 3)(x+ 2)gi 8 cos2 ®+ 2cos®¡ 1

= (4 cos®¡ 1)(2 cos®+ 1)

f8x2 + 2x¡ 1 = (4x¡ 1)(2x+ 1)g

5 a1¡ cos2 ®

1¡ cos®

=(1 + cos®)(1¡ cos®)

1¡ cos®

= 1 + cos®

bsin2 µ ¡ 1

sin µ + 1

=(sin µ + 1)(sin µ ¡ 1)

sin µ + 1

= sin µ ¡ 11 1

g sin2 µ ¡ 1

= ¡(1¡ sin2 µ)

= ¡ cos2 µ

fcos2 µ + sin2 µ = 1g

h 3¡ 3 sin2 µ

= 3(1¡ sin2 µ)

= 3 cos2 µ

fcos2 µ + sin2 µ = 1g

i 6 cos2 µ ¡ 6

= ¡6(1¡ cos2 µ)

= ¡6 sin2 µ

fcos2 µ + sin2 µ = 1g

j1¡ cos2 µ

sin2 µ

=sin2 µ

sin2 µ

fcos2 µ + sin2 µ = 1g= 1

k2¡ 2 cos2 µ

sin µ

=2(1¡ cos2 µ)

sin µ

=2 sin2 µ

sin µ

fcos2 µ + sin2 µ = 1g= 2 sin µ

lcos2 µ ¡ 1

sin µ

= ¡1¡ cos2 µ

sin µ

= ¡ sin2 µ

sin µ

fcos2 µ + sin2 µ = 1g= ¡ sin µ

1 1

230 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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ccos®¡ sin®

cos2 ®¡ sin2 ®

=cos®¡ sin®

(cos®+ sin®)(cos®¡ sin®)

=1

cos®+ sin®

dcos2 µ ¡ sin2 µ

cos µ + sin µ

=(cos µ + sin µ)(cos µ ¡ sin µ)

cos µ + sin µ

= cos µ ¡ sin µ

esinÁ+ cosÁ

cos2 Á¡ sin2 Á

=cosÁ+ sinÁ

(cosÁ+ sinÁ)(cosÁ¡ sinÁ)

=1

cosÁ¡ sinÁ

f4¡ 4 sin2 µ

2 cos µ

=4(1¡ sin2 µ)

2 cos µ

=4 cos2 µ

2 cos µ

= 2 cos µ

6 a (cos µ + sin µ)2 ¡ (cos µ ¡ sin µ)2

= cos2 µ + 2cos µ sin µ + sin2 µ ¡ (cos2 µ ¡ 2 cos µ sin µ + sin2 µ)

= cos2 µ + 2cos µ sin µ + sin2 µ ¡ cos2 µ + 2 cos µ sin µ ¡ sin2 µ

= 4 cos µ sin µ

b (4 sin µ + 3cos µ)2 + (3 sin µ ¡ 4 cos µ)2

= 16 sin2 µ + 24 sin µ cos µ + 9cos2 µ + 9 sin2 µ ¡ 24 sin µ cos µ + 16 cos2 µ

= 25 sin2 µ + 25 cos2 µ

= 25(sin2 µ + cos2 µ)

= 25

c (1¡ sin µ)

³1 +

1

sin µ

´= 1 +

1

sin µ¡ sin µ ¡ sin µ

sin µ

= 1 +1

sin µ¡ sin µ ¡ 1

=1

sin µ¡ sin µ

=1¡ sin2 µ

sin µ

=cos2 µ

sin µ

d

³1 +

1

cos µ

´(cos µ ¡ cos2 µ)

= cos µ ¡ cos2 µ + 1¡ cos µ

= 1¡ cos2 µ

= sin2 µ

ecos µ

1 + sin µ+

1 + sin µ

cos µ

=cos2 µ + (1 + sin µ)2

(1 + sin µ) cos µ

=cos2 µ + 1 + 2 sin µ + sin2 µ

(1 + sin µ) cos µ

=1 + 1 + 2 sin µ

(1 + sin µ) cos µ

=2 + 2 sin µ

(1 + sin µ) cos µ

=2(1 + sin µ)

(1 + sin µ) cos µ

=2

cos µ

fcos µ

1¡ sin µ¡ cos µ

1 + sin µ

=cos µ(1 + sin µ)¡ cos µ(1¡ sin µ)

(1¡ sin µ)(1 + sin µ)

=cos µ + cos µ sin µ ¡ cos µ + cos µ sin µ

1¡ sin2 µ

=2 cos µ sin µ

cos2 µ

=2 sin µ

cos µ

= 2 tan µ

1

1

1

1

1

1

1

1

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 231

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REVIEW SET 10A

1 sin µ =OPP

HYP=

5

13

cos µ =ADJ

HYP=

12

13

tan µ =OPP

ADJ=

5

12

2 a cos 64± =ADJ

HYP=

x

32

) x = 32 cos 64±

) x ¼ 14:0

b sinx =OPP

HYP=

132

229

) x = sin¡1(132

229)

) x ¼ 35:2±

4 Let the height be h m.

tan 34± =OPP

ADJ=

h

120

) h = 120 tan 34±

) h ¼ 80:9

The building is about 80:9 m high.

5 cos µ = ¡0:5781

) µ = cos¡1(¡0:5781)

) µ ¼ 125±

6 Suppose the ship is d km north of its starting point.

Now cos 56± =ADJ

HYP=

d

40

) d = 40 cos 56±

) d ¼ 22:37

) the ship is about 22:4 km north of its starting point.

7 a The required angle is BbGF.

tan µ =OPP

ADJ=

6

4=

3

2

) µ = tan¡1( 3

2)

) µ ¼ 56:3±

) the angle is about 56:3±.

5 cm 13 cm

12 cm

q

OPP

HYP

ADJ

34°

120 m

h m

y

x

-0.5781

d km

N

56°40 km

qG F

C B

6 cm

4 cm

3 µ = 90± ¡ 54±

) µ = 36±tan 54± =

OPP

ADJ=

17

x

) x =17

tan 54±

) x ¼ 12:4

sin 54± =OPP

HYP=

17

y

) y =17

sin 54±

) y ¼ 21:0

232 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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b The required angle is AbGE.

(EG)2 = 42 + 82 fPythagorasg) EG =

p80

tan µ =OPP

ADJ=

6p80

) µ = tan¡1( 6p80

)

) µ ¼ 33:9±

) the angle is about 33:9±.

8 Area = 1

2ab sinC

= 1

2£ 2:8£ 4:6£ sin 37±

¼ 3:88 km2

9 a cos µ = a

b sin µ = b

c tan µ =b

a

d sin(180± ¡ µ) = b

e cos(180± ¡ µ) = ¡a

f tan(180± ¡ µ) = ¡ b

a

10

H G

D

A B

F

C

E

8 cm4 cm

6 cm

q

-1 1x

y

P(a, b)(-a, b)

q

180° - q

P

A

B

d km

53°

16 2. km

18.9 km

Let the distance between them be d km.

AbPB is acute, so AbPB = 53±.

Using the cosine rule, a2 = b2 + c2 ¡ 2bc cosA

) d =p

16:22 + 18:92 ¡ 2£ 16:2£ 18:9£ cos 53±

) d ¼ 15:8

) the cyclists are about 15:8 km apart.

11 a Let [BD] be x m long.

Using the sine rule,x

sin 68± =197

sin 41±

) x =197 sin 68±

sin 41±

) x ¼ 278:413

So, [BD] is about 278 m long.

b AbBD = 180± ¡ 68± ¡ 41± = 71±

Area of quadrilateral ABCD

= area of 4ABD + area of 4BCD

¼ 1

2£ 197£ 278:413£ sin 71±

+ 1

2£ 207£ 278:413£ sin 46±

¼ 46 658 m2

¼ 4:67 ha

12 acos2 µ ¡ 1

cos µ + 1

=(cos µ + 1)(cos µ ¡ 1)

cos µ + 1

fa2 ¡ b2 = (a+ b)(a¡ b)g= cos µ ¡ 1

b4 cos µ

2 sin2 µ + 2cos2 µ

=4cos µ

2(sin2 µ + cos2 µ)

=4 cos µ

2

= 2 cos µ

1

1

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 233

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13sin µ

1¡ cos µ+

1¡ cos µ

sin µ

=sin2 µ + (1¡ cos µ)2

(1¡ cos µ) sin µ

=sin2 µ + 1¡ 2 cos µ + cos2 µ

(1¡ cos µ) sin µ

=1 + 1¡ 2 cos µ

(1¡ cos µ) sin µ

=2¡ 2 cos µ

(1¡ cos µ) sin µ

=2(1¡ cos µ)

(1¡ cos µ) sin µ

=2

sin µ

REVIEW SET 10B

1 a cos 74± ¼ 0:2756 b sin 132± ¼ 0:7431 c tan 97± ¼ ¡8:1443

2 a sinx =OPP

HYP=

5

8

) x = sin¡1(5

8)

) x ¼ 38:7±

b cosx =ADJ

HYP=

7:5

9:4

) x = cos¡1(7:5

9:4)

) x ¼ 37:1±

3 x2 + 192 = 322 fPythagorasg) x2 = 663

) x =p663 fas x > 0g

) x ¼ 25:7

sin® =OPP

HYP=

19

32

) ® = sin¡1( 19

32)

) ® ¼ 36:4±

µ = 90± ¡ ®

) µ ¼ 53:6±

4 Suppose the cliff is h km high.

tan 17:7± =OPP

ADJ=

h

2

) h = 2 tan 17:7±

) h ¼ 0:638

The cliff is about 638 m high.

5 Let the angle be µ.

cos µ =ADJ

HYP=

11

13

) µ = cos¡1(11

13)

) µ ¼ 32:2±

The angle is about 32:2±.

h kmcliff

2 km

17 7. °

13 cm

11 cm

q

234 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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6 cos2 µ + sin2 µ = 1

) cos2 µ + (3

4)2 = 1

) cos2 µ + 9

16= 1

) cos2 µ = 7

16

Since µ is obtuse, cos µ < 0

) cos µ = ¡p

7

4

7 a µ1 = 181± ¡ 180± = 1±

µ2 = µ1 falternate anglesg) µ2 = 1±

) QbRP = 360± ¡ 274± + 1±

= 87±

b The required angle is Á.

Let QbPR = ®.

Using the sine rule,sin®

60=

sinQbRP

d

) sin® ¼ 60£ sin 87±

76:066

) ® ¼ sin¡1

³60£ sin 87±

76:066

´) ® ¼ 51:97±

Á = ®+ µ1 falternate anglesg) Á ¼ 52:97±

) the ship must sail on the bearing 053:0±.

8 a BD2 = 202 + 202 fPythagorasg) BD = 20

p2 cm

) MD = 10p2 cm

cos µ =ADJ

HYP=

10p2

20=

p2

2=

1p2

) µ = 45±

) AbDM = 45±

b Triangle ACD is equilateral, so AbCD = 60±.

N

NN

181°

274°

f

q1

q2

50 kmd km

60 kmQ

P

a

R

M

B C

E D

20 cm 10~`2 cm

20 cm

A

M Dq

y

x

Er

Let PQ = d km.

Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) d =p

502 + 602 ¡ 2£ 50£ 60£ cos 87±

) d ¼ 76:066

It is about 76:1 km from P to Q.

Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY 235

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9

[AB] is about 275 m long.

b BbAC = 180± ¡ 54± ¡ 67± = 59±

) the area ¼ 1

2£ 242£ 275:349£ sin 59±

¼ 28 558 m2

¼ 2:86 ha

10 Using the cosine rule,

a2 = b2 + c2 ¡ 2bc cosA

) 102 = 122 + x2 ¡ 2£ 12£ x£ cos 40±

) 100 = 144 + x2 ¡ 24 cos 40±x

) x2 ¡ 24 cos 40±x+ 44 = 0

Using the quadratic formula,

x =¡(¡24 cos 40±)§

p(24 cos 40±)2 ¡ 4£ 1£ 44

2£ 1

) x = 12 cos 40± § 1

2

p576 cos2 40± ¡ 176

) x ¼ 9:1925§ 6:3642

) x ¼ 15:56 or 2:83

11 asin µ

cos µ= tan µ b

=(1 + sin µ)(1¡ sin µ)

1 + sin µfa2 ¡ b2 = (a+ b)(a¡ b)g

= 1¡ sin µc

3¡ 3 cos2 µ

sin µ

=3(1¡ cos2 µ)

sin µ

=3 sin2 µ

sin µf cos2 µ + sin2 µ = 1g

= 3 sin µ

12 a (sin µ + 2cos µ)2 + (2 sin µ ¡ cos µ)2

= sin2 µ + 4 sin µ cos µ + 4cos2 µ + 4 sin2 µ ¡ 4 sin µ cos µ + cos2 µ

= 5 sin2 µ + 5cos2 µ

= 5(sin2 µ + cos2 µ)

= 5

b (1 + sin µ)

³1¡ 1

sin µ

´= 1¡ 1

sin µ+ sin µ ¡ 1

= ¡ 1

sin µ+ sin µ

= ¡1¡ sin2 µ

sin µ

= ¡cos2 µ

sin µ

A

C

B

x m242 m

54°

67°

A

B C

40°

12 m x m

10 m

a Let [AB] be x m long.

Using the sine rule,x

sin 67± =242

sin 54±

) x =242 sin 67±

sin 54±

) x ¼ 275:349

1¡ sin2 µ

1 + sin µ

1

1

236 Mathematics MYP5+ (2nd edn), Chapter 10 – TRIGONOMETRY

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