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CHAPTER 38 INTRODUCTION TO TRIGONOMETRY

EXERCISE 158 Page 427

1. Find the length of side x.

By Pythagoras’s theorem, 2 2 241 40x= + from which, 2 2 241 40x = − and x = 2 241 40− = 9 cm 2. Find the length of side x.

By Pythagoras’s theorem, 2 2 225 7x= + from which, 2 2 225 7x = − and x = 2 225 7− = 24 m 3. Find the length of side x, correct to 3 significant figures.

By Pythagoras’s theorem, 2 2 24.7 8.3x = + from which, x = 2 24.7 8.3+ = 9.54 mm 4. In a triangle ABC, AB = 17 cm, BC = 12 cm and ∠ABC = 90°. Determine the length of AC, correct to 2 decimal places.

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Triangle ABC is shown sketched below.

By Pythagoras’s theorem, 2 2 217 12AC = + from which, AC = 2 217 12+ = 20.81cm 5. A tent peg is 4.0 m away from a 6.0 m high tent. What length of rope, correct to the nearest centimetre, runs from the top of the tent to the peg? The tent peg is shown as C in the sketch below, with AB being the tent height.

By Pythagoras, length of rope, AC = 2 26.0 4.0+ = 7.21 m 6. In a triangle ABC, ∠B is a right-angle, AB = 6.92 cm and BC = 8.78 cm. Find the length of the hypotenuse. Triangle ABC is shown sketched below.

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By Pythagoras’s theorem, 2 2 26.92 8.78AC = + from which, hypotenuse, AC = 2 26.92 8.78+ = 11.18 cm 7. In a triangle CDE, D = 90°, CD = 14.83 mm and CE = 28.31 mm. Determine the length of DE. Triangle CDE is shown sketched below.

By Pythagoras’s theorem, 2 2 228.31 14.83DE= + from which, 2 2 228.31 14.83DE = − and DE = 2 228.31 14.83− = 24.11 mm 8. Show that if a triangle has sides of 8, 15 and 17 cm it is right-angled. Pythagoras’s theorem applies to right-angled triangles only Assuming the hypotenuse is 17, then 2 2 217 15 8= + i.e. 289 = 225 + 64 Since Pythagoras’s theorem may be applied, the triangle must be right-angled 9. Triangle PQR is isosceles, Q being a right angle. If the hypotenuse is 38.46 cm find (a) the lengths

of sides PQ and QR, and (b) the value of ∠QPR.

(a) Since triangle PQR in the diagram below is isosceles, PQ = QR From Pythagoras, 2 2 2 2(38.47) ( ) ( ) 2( )PQ QR PQ= + =

from which, ( )22 38.47

2PQ = and PQ =

238.47 38.472 2

= = 27.20 cm

Hence, PQ = QR = 27.20 cm

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(b) Since triangle PQR is isosceles, ∠P = ∠R and since ∠Q = 90°, then ∠P + ∠R = 90° Hence, ∠QPR = 45° (=∠QRP) 10. A man cycles 24 km due south and then 20 km due east. Another man, starting at the same time

as the first man, cycles 32 km due east and then 7 km due south. Find the distance between the

two men.

With reference to the diagram below, AB = 32 – 20 = 12 km and BC = 24 – 7 = 17 km

Hence, distance between the two men, AC = ( )2 212 17+ = 20.81 km by Pythagoras

11. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall. How

far up the wall (to the nearest centimetre) does the ladder reach? If the foot of the ladder is now

moved 30 cm further away from the wall, how far does the top of the ladder fall?

Distance up the wall, AB = ( )2 23.5 1.0− = 3.35 m by Pythagoras

( ) ( ) ( )2 2 2 2' ' ' ' 3.5 1.30 3.25A B A C BC m = − = − =

Hence, the amount the top of the ladder has moved down the wall, given by AA′ = 3.35 – 3.25 = 0.10 m or 10 cm

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12. Two ships leave a port at the same time. One travels due west at 18.4 knots and the other due

south at 27.6 knots. If 1 knot = 1 nautical mile per hour, calculate how far apart the two ships

are after four hours.

After four hours, the ship travelling west travels 4 × 18.4 = 73.6 km, and the ship travelling south travels 4 × 27.6 = 110.4 km, as shown in the diagram below

Hence, distance apart after four hours = ( )2 273.6 110.4+ = 132.7 km by Pythagoras

13. The diagram shows a bolt rounded off at one end. Determine the dimension h.

Part of the bolt is shown below

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From Pythagoras, 2 2 2AB AC BC= + i.e. 2 2 245 16 BC= + and 2 2 245 16BC = − from which, length BC = ( )2 245 16 1769− = = 42.06 mm Length BD = radius = 45 mm, hence, h = CD = BD – BC = 45 – 42.06 = 2.94 mm 14. The diagram shows a cross-section of a component that is to be made from a round bar. If the

diameter of the bar is 74 mm, calculate the dimension x.

From the above diagram, 2 2 20 0B A AB= +

where AB = 2x , 0B = 37 mm (radius) and 0A = 72 – 37 = 35 mm

Hence, 2 2 2 2 20 0 37 35AB B A= − = − and AB = ( )2 237 35− = 12 mm

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Hence dimension, x = 2 × 12 = 24 mm

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EXERCISE 159 Page 430

1. Sketch a triangle XYZ such that ∠Y = 90°, XY = 9 cm and YZ = 40 cm. Determine sin Z, cos Z,

tan X and cos X.

Triangle XYZ is shown sketched below.

By Pythagoras’s theorem, XZ = 2 240 9+ = 41

sin Z = oppositehypotenuse

XYXZ

= = 941

cos Z = adjacenthypotenuse

YZXZ

= = 4041

tan X = oppositeadjacent

YZXY

= =409

cos X = adjacenthypotenuse

XYXZ

= = 941

2. In triangle ABC shown below, find sin A, cos A, tan A, sin B, cos B and tan B.

By Pythagoras’s theorem, AC = 2 25 3− = 4

sin A = oppositehypotenuse

BCAB

= = 35

cos A = adjacenthypotenuse

ACAB

= = 45

tan A = oppositeadjacent

BCAC

= =34

sin B = oppositehypotenuse

ACAB

= = 45

cos B = adjacenthypotenuse

BCAB

= = 35

tan B = oppositeadjacent

ACBC

= = 43

3. If cos A = 1517

find sin A and tan A, in fraction form.

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Triangle ABC is shown sketched below where cos A = 1517

By Pythagoras’s theorem, BC = 2 217 15− = 8

sin A = oppositehypotenuse

BCAC

= = 817

and tan A = oppositeadjacent

BCAB

= =8

15

4. If tan X = 15112

, find sin X and cos X, in fraction form.

Triangle XYZ is shown sketched below where tan X = 15112

By Pythagoras’s theorem, XZ = 2 215 112+ = 113

sin X = oppositehypotenuse

YZXZ

= = 15113

and cos X = adjacenthypotenuse

XYXZ

= =112113

5. For the right-angled triangle shown, find: (a) sin α (b) cos θ (c) tan θ

(a) sin α = oppositehypotenuse

= 1517

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(b) cos θ = adjacenthypotenuse

= 1517

(c) tan θ = oppositeadjacent

= 815

6. If tan θ = 724

, find sin θ and cos θ in fraction form.

Triangle ABC is shown sketched below where tan θ = 724

By Pythagoras’s theorem, AC = 2 224 7+ = 25

sin θ = oppositehypotenuse

ABAC

= = 725

and cos θ = adjacenthypotenuse

BCAC

= =2425

7. Point P lies at coordinate (–3, 1) and point Q at (5, –4). Determine (a) the distance PQ, and (b) the gradient of the straight line PQ. (a) From the diagram below, PQ = ( )2 25 8+ = 9.434 by Pythagoras

(b) Gradient of PQ = 1 4 5

3 5 8− −

=− − −

= –0.625

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EXERCISE 160 Page 432

1. Determine, correct to 4 decimal places, 3 sin 66° 41′ Using a calculator, 3 sin 66° 41′ = 2.7550, correct to 4 decimal places 2. Determine, correct to 3 decimal places, 5 cos 14° 15′ Using a calculator, 5 cos 14° 15′ = 4.846, correct to 3 decimal places 3. Determine, correct to 4 significant figures, 7 tan 79° 9′

Using a calculator, 7 tan 79° 9′ = 36.52, correct to 4 significant figures

4. Determine (a) sine 23π (b) cos 1.681 (c) tan 3.672

Note that with no degrees sign, these angles are in radians

(a) Using a calculator, sine 23π

= 0.8660

(b) Using a calculator, cos 1.681 = –0.1010

(c) Using a calculator, tan 3.672 = 0.5865

5. Find the acute angle 1sin 0.6734− in degrees, correct to 2 decimal places. Using a calculator, 1sin 0.6734− = 42.33° 6. Find the acute angle 1cos 0.9648− in degrees, correct to 2 decimal places. Using a calculator, 1cos 0.9648− = 15.25° 7. Find the acute angle 1tan 3.4385− in degrees, correct to 2 decimal places.

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Using a calculator, 1tan 3.4385− = 73.78° 8. Find the acute angle 1sin 0.1381− in degrees and minutes. Using a calculator, 1sin 0.1381− = 7.9379...° = 7° 56´ correct to the nearest minute 9. Find the acute angle 1cos 0.8539− in degrees and minutes. Using a calculator, 1cos 0.8539− = 31.36157...° = 31° 22´ correct to the nearest minute 10. Find the acute angle 1tan 0.8971− in degrees and minutes. Using a calculator, 1tan 0.8971− = 41.89528...° = 41° 54´ correct to the nearest minute 11. In the triangle shown, determine angle θ, correct to 2 decimal places.

From trigonometric ratios, tan θ = 59

from which, θ = 1 5tan9

−

= 29.05°

12. In the triangle shown, determine angle θ in degrees and minutes.

From trigonometric ratios, sin θ = 823

from which, θ = 1 8sin23

−

= 20.35° = 20° 21′

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13. Evaluate, correct to 4 decimal places: 4.5 cos 67 34 ' sin 902 tan 45

° − °°

Using a calculator, 4.5 cos 67 34 ' sin 90

2 tan 45° − °

° = 0.3586, correct to 4 decimal places

14. Evaluate, correct to 4 significant figures: ( )( )3sin 37.83 2.5 tan 57.484.1 cos 12.56

° °°

Using a calculator, ( )( )3sin 37.83 2.5 tan 57.48

4.1 cos 12.56° °

° = 1.803, correct to 4 significant figures

15. For the supported beam AB shown in the diagram, determine (a) the angle the supporting stay

CD makes with the beam, i.e. θ, correct to the nearest degree, (b) the length of the stay, CD,

correct to the nearest centimetre.

(a) tan θ = 4.365.20

ACAD

= hence angle θ = 1 4.36tan5.20

−

= 39.98° = 40° correct to nearest degree

(b) By Pythagoras, 2 2 24.36 5.20CD = +

from which, CD = 2 24.36 5.20+ = 6.79 m

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EXERCISE 161 Page 434

1. If cos X = 725

determine the value of the other five trigonometric ratios.

A right-angled triangle XYZ is shown below.

Since cos X = 725

, then XY = 7 units and XZ = 25 units

Using Pythagoras’s theorem: 252 = 72 + YZ2 from which YZ = 2 225 7− = 24 units

Thus, sin X = 2425

, tan X = 247

= 3 37

, cosec X = 2524

= 1 124

,

sec X = 257

= 3 47

and cot X = 724

2. If sin θ = 0.40 and cos θ = 0.50 determine the values of cosec θ, sec θ, tan θ and cot θ.

cosec θ = 1sinθ

= 10.40

= 2.50 sec θ = 1cosθ

= 10.50

= 2.00

tan θ = sincos

θθ

= 0.400.50

= 0.80 cot θ = cossin

θθ

= 0.500.40

= 1.25

3. Evaluate correct to 4 decimal places: (a) secant 73° (b) secant 286.45° (c) secant 155° 41’

(a) sec 73° = 1cos 73°

= 3.4203

(b) sec 286.45° = 1cos 286.45°

= 3.5313

(c) sec 155° 41' = 1cos155 41'°

= 141cos15560°

= –1.0974

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4. Evaluate correct to 4 decimal places:

(a) cosecant 213° (b) cosecant 15.62° (c) cosecant 311° 50'

(a) cosec 213° = 1sin 213°

= –1.8361

(b) cosec 15.62° = 1sin15.62°

= 3.7139

(c) cosec 311° 50' = 1sin 311 50 '°

= 150sin 31160°

= –1.3421

5. Evaluate correct to 4 decimal places:

(a) cotangent 71° (b) cotangent 151.62° (c) cotangent 321° 23'

(a) cot 71° = 1tan 71°

= 0.3443

(a) cot 151.62° = 1tan151.62°

= –1.8510

(b) cot 321°23' = 1tan 321 23'°

= 123tan 32160°

= –1.2519

6. Evaluate correct to 4 decimal places: (a) sec 8π (b) cosec 2.961 (c) cot 2.612

(a) sec 8π = 1

cos8π

= 1.0824

(b) cosec 2.961 = 1sin 2.961

= 5.5675

(c) cot 2.612 = 1tan 2.612

= –1.7083

7. Determine the acute angle of the following in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places): sec 1− 1.6214

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sec 1− 1.6214 = cos 1−1

1.6214

= cos 1− 0.61675... = 51.92° or 51°55' or 0.906 radians

8. Determine the acute angle of the following in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places): cosec 1− 2.4891

cosec 1− 2.4891 = sin 1−1

2.4891

= sin 1− 0.40175... = 23.69° or 23°41' or 0.413 radians

9. Determine the acute angle of the following in degrees (correct to 2 decimal places), degrees and

minutes, and in radians (correct to 3 decimal places): cot 1− 1.9614

cot 1− 1.9614 = tan 1−1

1.9614

= tan 1− 0.50983.. = 27.01° or 27°1' or 0.471 radians

10. Evaluate 6.4cosec 29 5' sec812cot12

° − °°

correct to 4 significant figures.

Using a calculator, 6.4cosec 29 5' sec812cot12

° − °°

= 0.7199, correct to 4 significant figures

11. If tan x = 1.5276, determine sec x, cosec x, and cot x (assume x is an acute angle).

If tan x = 1.5276 then x = 1tan 1.5276− = 56.79°

Hence, sec 56.79° = 1cos56.79°

= 1.8258

cosec 56.79° = 1sin 56.79°

= 1.1952

and cot 56.79° = 1tan 56.79°

= 0.6546

12. Evaluate correct to 4 significant figures: 3 cot 14° 15' sec 23° 9'

Using a calculator, 3 cot 14° 15' sec 23° 9' = 12.85, correct to 4 significant figures

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13. Evaluate correct to 4 significant figures: cosec 27 19 ' sec 45 29 '1 cosec 27 19 'sec 45 29 '

° + °− ° °

Using a calculator, cosec 27 19 ' sec 45 29 '1 cosec 27 19 'sec 45 29 '

° + °− ° °

= –1.710, correct to 4 significant figures

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EXERCISE 162 Page 436

1. Evaluate, without using a calculator: 3 sin 30° – 2 cos 60°

3 sin 30° – 2 cos 60° = 1 1 33 2 12 2 2

− = −

= 12

2. Evaluate, without using a calculator, leaving in surd form: 5 tan 60° – 3 sin 60°

5 tan 60° – 3 sin 60° = ( ) 3 35 3 3 5 3 32 2

− = −

= 7 3

2

3. Evaluate, without using a calculator: tan 603tan 30

°°

tan 60 3 3 3 3

3tan 30 3 3133

°= = =

°

= 1

4. Evaluate, without using a calculator, leaving in surd form: (tan 45°)(4 cos 60° – 2 sin 60°)

(tan 45°)(4 cos 60° – 2 sin 60°) = ( ) 1 31 4 22 2

− = 2 3−

5. Evaluate, without using a calculator, leaving in surd form: tan 60 tan 301 tan 30 tan 60

° − °+ ° °

( )

1 3 1 23tan 60 tan 30 3 3 3

1 tan 30 tan 60 1 1 211 33

−−

°− °= = =

+ ° ° + +

= 13

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EXERCISE 163 Page 437

1. Calculate the dimensions shown as x in (a) to (f), each correct to 4 significant figures.

(a) sin 70° = 13.0

x from which, x = 13.0 sin 70° = 12.22

(b) sin 22° = 15.0

x from which, x = 15.0 sin 22° = 5.619

(c) cos 29° = 17.0

x from which, x = 17.0 cos 29° = 14.87

(d) cos 59° = 4.30x

from which, x = 4.30cos59°

= 8.349

(e) tan 43° = 6.0x from which, x = 6.0 tan 43° = 5.595

(f) tan 53° = 7.0x

from which, x = 7.0tan 53°

= 5.275

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2. Find the unknown sides and angles in the right-angled triangles shown. The dimensions shown are in centimetres.

(a) By Pythagoras, AC = 2 23.0 5.0+ = 5.831 cm

tan C = 3.05.0

from which, ∠C = 1 3.0tan5.0

−

= 30.96°

and ∠A =180° – 90° – 30.96° = 59.04° (b) By Pythagoras, DE = 2 28.0 4.0− = 6.928 cm

sin D = 4.08.0

from which, ∠D = 1 4.0sin8.0

−

= 30°

and ∠F =180° – 90° – 30° = 60° (c) ∠J =180° – 90° – 28° = 62°

sin 28° = 12.0HJ from which, HJ = 12.0 sin 28° = 5.634 cm

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By Pythagoras, GH = 2 212.0 5.634− = 10.60 cm (d) ∠L =180° – 90° – 27° = 63°

sin 27° = 15.0LM from which, LM = 15.0 sin 27° = 6.810 cm

By Pythagoras, KM = 2 215.0 6.810− = 13.37 cm (e) ∠N =180° – 90° – 64° = 26°

tan 64° = 4.0NP from which, NP = 4.0 tan 64° = 8.201 cm

By Pythagoras, ON = 2 28.201 4.0+ = 9.124 cm (f) ∠S =180° – 90° – 41° = 49°

tan 41° = 5.0RS from which, RS = 5.0 tan 41° = 4.346 cm

By Pythagoras, QS = 2 24.346 5.0+ = 6.625 cm 3. A ladder rests against the top of the perpendicular wall of a building and makes an angle of 73°

with the ground. If the foot of the ladder is 2 m from the wall, calculate the height of the building.

The ladder is shown in the diagram below, where BC is the height of the building.

Tan 73° = 2

BC from which, height of building, BC = 2 tan 73° = 6.54 m

4. Determine the length x in the diagram.

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From triangle ABC in the sketch above,

tan 28° = 5BCAB x

= from which, x = 5tan 28°

= 9.40 mm

5. A symmetrical part of a bridge lattice is shown. If AB = 6 m, angle BAD = 56° and E is the

midpoint of ABCD, determine the height h, correct to the nearest centimetre.

In triangle ABE, ∠BAE = 562° = 28°

and sin 28° = 6

BE BEAB

= from which, BE = 6 sin 28° = 2.817 m

Hence, height, h = 2 × 2.817 = 5.63 m

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EXERCISE 164 Page 440

1. A vertical tower stands on level ground. At a point 105 m from the foot of the tower the angle of

elevation of the top is 19°. Find the height of the tower.

A side view is shown below with the tower being AB.

Tan 19° = 105AB from which, height of tower, AB = 105 tan 19° = 36.15 m

2. If the angle of elevation of the top of a vertical 30 m high aerial is 32°, how far is it to the aerial?

A side view is shown below with the aerial being AB.

Tan 32° = 30BC

from which, distance to aerial, BC = 30tan 32°

= 48 m

3. From the top of a vertical cliff 90.0 m high the angle of depression of a boat is 19° 50′. Determine

the distance of the boat from the cliff.

A side view is shown below with the cliff being AB. Since the angle of depression of a boat is 19° 50′ then ∠ACB = 19° 50'

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Tan 19° 50' = 90.0

BC from which, distance of boat to the cliff, BC = 90.0

tan19 50 '° = 249.5 m

4. From the top of a vertical cliff 80.0 m high the angles of depression of two buoys lying due west

of the cliff are 23° and 15°, respectively. How far are the buoys apart?

In the diagram below, the two buoys are shown as A and B

Tan 15° = 80

AC from which, AC = 80

tan15° = 298.56 m

Tan 23° = 80BC

from which, BC = 80tan 23°

= 188.47 m

Hence, distance apart, AB = AC – BC = 298.56 – 188.47 = 110.1 m 5. From a point on horizontal ground a surveyor measures the angle of elevation of the top of a

flagpole as 18° 40′. He moves 50 m nearer to the flagpole and measures the angle of elevation as

26° 22′. Determine the height of the flagpole.

A side view is shown below with the flagpole being AB.

Tan 18°40' = 50

hBD+

from which, height h = (tan 18°40´)(50 + BD)

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= (0.337833)(50 + BD) = 16.89165 + 0.337833(BD)

Tan 26°22' = hBD

from which, height, h = (tan 26°22´)(BD)

= (0.495679)(BD) (1) Equating the h values gives: 16.89165 + 0.337833(BD) = (0.495679)(BD) from which, 16.89165 = 0.495679(BD) – 0.337833(BD) i.e. 16.89165 = 0.157846(BD)

and BD = 16.891650.157846

= 107.01 m

Hence, from equation (1), height of flagpole = (0.495679)(BD) = 0.495679 × 107.01 = 53.0 m 6. A flagpole stands on the edge of the top of a building. At a point 200 m from the building the

angles of elevation of the top and bottom of the pole are 32° and 30°, respectively. Calculate the

height of the flagpole.

In the diagram below, the flagpole is shown as AB.

Tan 32° =

200AC from which, AC = 200 tan 32° = 124.97 m

Tan 30° = 200BC from which, BC = 200 tan 30° = 115.47 m

Hence, height of flagpole, AB = AC – BC = 124.97 – 115.47 = 9.50 m 7. From a ship at sea, the angles of elevation of the top and bottom of a vertical lighthouse standing

on the edge of a vertical cliff are 31° and 26°, respectively. If the lighthouse is 25.0 m high,

calculate the height of the cliff.

A side view is shown below with the lighthouse being AB.

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Tan 26° = hDC

from which, DC = tan 26

h°

= 2.0503h

Tan 31° = 25hDC+ from which, DC = 25

tan 31h +

° = 1.66428(h + 25) = 1.66428h + 41.607

Equating the DC values gives: 2.0503h = 1.66428h + 41.607

i.e. 2.0503h – 1.66428h = 41.607

i.e. 0.3860h = 41.607

from which, height of cliff, h = 41.6070.3860

= 107.8 m

8. From a window 4.2 m above horizontal ground the angle of depression of the foot of a building

across the road is 24° and the angle of elevation of the top of the building is 34°. Determine,

correct to the nearest centimetre, the width of the road and the height of the building.

In the diagram below, D is the window, the width of the road is AB and the height of the building across the road is BC.

In the triangle ABD, ∠D = 90° – 24° = 66°

Tan 66° = 4.2AB hence, width of road, AB = 4.2 tan 66° = 9.43 m

From triangle DEC, tan 34° = 9.43

CE CE CEDE AB

= = from which, CE = 9.43 tan 34° = 6.36 m

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Hence, height of building, BC = CE + EB = CE + AD = 6.36 + 4.2 = 10.56 m 9. The elevation of a tower from two points, one due west of the tower and the other due east of it

are 20° and 24°, respectively, and the two points of observation are 300 m apart. Find the height

of the tower to the nearest metre.

In the diagram below, the height of the tower is AB and the two observation points are at C and D.

Tan 20° = AB

BC from which, AB = BC tan 20°

Tan 24° = 300

ABBC−

from which, AB = (300 – BC) tan 24°

i.e. BC tan 20° = (300 – BC) tan 24° = 300 tan 24° – BC tan 24° i.e. 0.36397 BC = 133.57 – 0.44523 BC i.e. 0.8092 BC = 133.57

and BC = 133.570.8092

= 165.06 m

Tan 20° = 165.06

AB from which, height of tower, AB = 165.06 tan 20° = 60 m, to the nearest metre