CHAPTER 38 INTRODUCTION TO TRIGONOMETRY
EXERCISE 158 Page 427
1. Find the length of side x.
By Pythagoras’s theorem,
from which,
2
41=
x 2 + 402
=
x 2 412 − 402
and x =
412 − 402 = 9 cm
2. Find the length of side x.
By Pythagoras’s theorem,
from which,
2
25=
x2 + 72
=
x 2 252 − 7 2
and x =
252 − 7 2 = 24 m
3. Find the length of side x, correct to 3 significant figures.
By Pythagoras’s theorem,
from which,
=
x 2 4.7 2 + 8.32
x=
4.7 2 + 8.32 = 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,
from which,
2
AC=
17 2 + 122
AC = 17 2 + 122 = 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 =
6.02 + 4.02 = 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,
=
AC 2 6.922 + 8.782
6.922 + 8.782 = 11.18 cm
from which, hypotenuse, AC =
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,
from which,
2
28.31
=
DE 2 + 14.832
=
DE 2 28.312 − 14.832
and DE =
28.312 − 14.832 = 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
2
Assuming the hypotenuse is 17, then 17=
152 + 82
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, (38.47) 2 = ( PQ) 2 + (QR) 2 = 2( PQ) 2
from which,
Hence,
( PQ )
2
=
38.47 2
2
and
PQ =
38.47 2 38.47
= 27.20 cm
=
2
2
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 =
(122 + 172 )
= 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 =
A' B =
( 3.52 − 1.02 )
( A ' C ')2 − ( BC ')2  =


= 3.35 m by Pythagoras
( 3.52 − 1.302 ) = 3.25 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 =
( 73.62 + 110.42 )
= 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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2
From Pythagoras, AB
=
AC 2 + BC 2
from which, length BC =
i.e.
2
45
=
162 + BC 2 and
1769
( 452 − 162 ) =
2
BC
=
452 − 162
= 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,
where AB =
Hence,
0=
B 2 0 A2 + AB 2
x
, 0B = 37 mm (radius) and 0A = 72 – 37 = 35 mm
2
AB 2 =0 B 2 − 0 A2 =37 2 − 352
and
639
AB =
( 372 − 352 )
= 12 mm
© 2014, John Bird
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 =
402 + 92 = 41
sin Z =
XY
opposite
9
=
=
hypotenuse XZ
41
cos Z =
adjacent
YZ 40
=
=
hypotenuse XZ 41
tan X =
opposite YZ 40
= =
adjacent XY
9
cos X =
adjacent
XY
9
=
=
hypotenuse XZ 41
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 =
sin A =
opposite
BC
3
=
=
hypotenuse AB
5
sin B =
opposite
AC
4
=
=
hypotenuse AB
5
3. If cos A =
52 − 32 = 4
cos A =
cos B =
adjacent
AC
4
=
=
hypotenuse AB
5
adjacent
BC
3
=
=
hypotenuse AB
5
tan A =
opposite BC 3
= =
adjacent AC 4
tan B =
opposite AC
4
=
=
adjacent BC
3
15
find sin A and tan A, in fraction form.
17
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Triangle ABC is shown sketched below where cos A =
15
17
By Pythagoras’s theorem, BC = 17 2 − 152 = 8
sin A =
opposite
BC
8
=
=
hypotenuse AC 17
4. If tan X =
and
tan A =
opposite BC 8
= =
adjacent AB 15
15
, find sin X and cos X, in fraction form.
112
Triangle XYZ is shown sketched below where tan X =
15
112
By Pythagoras’s theorem, XZ = 152 + 1122 = 113
sin X =
opposite
YZ
15
=
=
hypotenuse XZ
113
and
cos X =
adjacent
XY 112
= =
hypotenuse XZ 113
5. For the right-angled triangle shown, find: (a) sin α (b) cos θ (c) tan θ
(a) sin α =
opposite
15
=
hypotenuse 17
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(b) cos θ =
adjacent
15
=
hypotenuse 17
(c) tan θ =
opposite
8
=
adjacent 15
6. If tan θ =
7
, find sin θ and cos θ in fraction form.
24
Triangle ABC is shown sketched below where tan θ =
By Pythagoras’s theorem, AC =
sin θ =
AB
opposite
7
=
=
hypotenuse AC
25
7
24
242 + 7 2 = 25
and
cos θ =
BC 24
adjacent
= =
hypotenuse AC 25
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 =
(b) Gradient of PQ =
( 52 + 82 )
= 9.434 by Pythagoras
1 − −4 5
= –0.625
=
−3 − 5 −8
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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
2π
3
(b) cos 1.681
(c) tan 3.672
Note that with no degrees sign, these angles are in radians
(a) Using a calculator, sine
2π
= 0.8660
3
(b) Using a calculator, cos 1.681 = –0.1010
(c) Using a calculator, tan 3.672 = 0.5865
5. Find the acute angle sin −1 0.6734 in degrees, correct to 2 decimal places.
Using a calculator, sin −1 0.6734 = 42.33°
6. Find the acute angle cos −1 0.9648 in degrees, correct to 2 decimal places.
Using a calculator, cos −1 0.9648 = 15.25°
7. Find the acute angle tan −1 3.4385 in degrees, correct to 2 decimal places.
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Using a calculator, tan −1 3.4385 = 73.78°
8. Find the acute angle sin −1 0.1381 in degrees and minutes.
Using a calculator, sin −1 0.1381 = 7.9379...° = 7° 56´ correct to the nearest minute
9. Find the acute angle cos −1 0.8539 in degrees and minutes.
Using a calculator, cos −1 0.8539 = 31.36157...° = 31° 22´ correct to the nearest minute
10. Find the acute angle tan −1 0.8971 in degrees and minutes.
Using a calculator, tan −1 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 =
5
from which,
9
5
θ = tan −1   = 29.05°
9
12. In the triangle shown, determine angle θ in degrees and minutes.
From trigonometric ratios, sin θ =
8
from which,
23
645
 8 
θ = sin −1   = 20.35° = 20° 21′
 23 
© 2014, John Bird
13. Evaluate, correct to 4 decimal places:
Using a calculator,
4.5 cos 67° 34 '− sin 90°
2 tan 45°
4.5 cos 67° 34 '− sin 90°
= 0.3586, correct to 4 decimal places
2 tan 45°
14. Evaluate, correct to 4 significant figures:
Using a calculator,
( 3sin 37.83° )( 2.5 tan 57.48° )
4.1 cos 12.56°
( 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 =
AC 4.36
=
AD 5.20
 4.36 
hence angle θ = tan −1 
 = 39.98° = 40° correct to nearest degree
 5.20 
2
(b) By Pythagoras, CD
=
4.362 + 5.202
from which,
CD =
4.362 + 5.202 = 6.79 m
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EXERCISE 161 Page 434
1. If cos X =
7
determine the value of the other five trigonometric ratios.
25
A right-angled triangle XYZ is shown below.
7
, then XY = 7 units and XZ = 25 units
25
Since cos X =
Using Pythagoras’s theorem: 252 = 72 + YZ2 from which YZ =
Thus, sin X =
24
,
25
sec X =
tan X =
25
4
=3
7
7
24
3
=3 ,
7
7
and
cosec X =
cot X =
252 − 7 2 = 24 units
25
1
=1 ,
24
24
7
24
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 =
tan θ =
1
1
=
= 2.50
sin θ
0.40
0.40
sin θ
=
= 0.80
0.50
cos θ
sec θ =
1
1
=
= 2.00
cos θ
0.50
cot θ =
0.50
cos θ
=
= 1.25
0.40
sin θ
3. Evaluate correct to 4 decimal places: (a) secant 73° (b) secant 286.45° (c) secant 155° 41’
(a) sec 73° =
1
= 3.4203
cos 73°
(b) sec 286.45° =
(c) sec 155° 41' =
1
= 3.5313
cos 286.45°
1
1
=
= –1.0974
cos155° 41' cos155 41°
60
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© 2014, John Bird
4. Evaluate correct to 4 decimal places:
(a) cosecant 213° (b) cosecant 15.62° (c) cosecant 311° 50'
1
= –1.8361
sin 213°
(a) cosec 213° =
1
= 3.7139
sin15.62°
1
1
(c) cosec 311° 50' =
=
= –1.3421
sin 311°50 ' sin 311 50°
60
(b) cosec 15.62° =
5. Evaluate correct to 4 decimal places:
(a) cotangent 71° (b) cotangent 151.62° (c) cotangent 321° 23'
(a) cot 71° =
1
= 0.3443
tan 71°
1
= –1.8510
tan151.62°
(a) cot 151.62° =
1
1
=
= –1.2519
23°
tan 321°23'
tan 321
60
(b) cot 321°23' =
6. Evaluate correct to 4 decimal places: (a) sec
(a) sec
π
8
1
=
cos
π
8
(b) cosec 2.961 (c) cot 2.612
= 1.0824
8
(b) cosec 2.961 =
(c) cot 2.612 =
π
1
= 5.5675
sin 2.961
1
= –1.7083
tan 2.612
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):
648
sec −1 1.6214
© 2014, John Bird
 1 
−1
sec −1 1.6214 = cos −1 
 = cos 0.61675... = 51.92° or 51°55' or 0.906 radians
 1.6214 
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
 1 
−1
cosec −1 2.4891 = sin −1 
 = sin 0.40175... = 23.69° or 23°41' or 0.413 radians
 2.4891 
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
 1 
−1
cot −1 1.9614 = tan −1 
 = tan 0.50983.. = 27.01° or 27°1' or 0.471 radians
 1.9614 
10. Evaluate
6.4cosec 29°5' − sec81°
correct to 4 significant figures.
2 cot12°
Using a calculator,
6.4cosec 29°5' − sec81°
= 0.7199, correct to 4 significant figures
2 cot12°
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 = tan −1 1.5276 = 56.79°
Hence, sec 56.79° =
1
= 1.8258
cos 56.79°
cosec 56.79° =
and
cot 56.79° =
1
= 1.1952
sin 56.79°
1
= 0.6546
tan 56.79°
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:
Using a calculator,
cosec 27°19 ' + sec 45°29 '
1 − cosec 27°19 'sec 45°29 '
cosec 27°19 ' + sec 45°29 '
= –1.710, correct to 4 significant figures
1 − cosec 27°19 'sec 45°29 '
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EXERCISE 162 Page 436
1. Evaluate, without using a calculator: 3 sin 30° – 2 cos 60°
1
1 1 3
−1 =
3 sin 30° – 2 cos 60° = 3   − 2   =
2
2 2 2
2. Evaluate, without using a calculator, leaving in surd form: 5 tan 60° – 3 sin 60°
5 tan 60° – 3 sin 60° = 5
 3
3
7
3
5 3−
3 =
 =
2
2
 2 
( 3 ) − 3 
3. Evaluate, without using a calculator:
tan 60°
=
3 tan 30°
3
=
 1 
3

 3
tan 60°
3 tan 30°
3 3 3
=1
=
3
3
4. Evaluate, without using a calculator, leaving in surd form: (tan 45°)(4 cos 60° – 2 sin 60°)
  1   3 
(tan 45°)(4 cos 60° – 2 sin 60°) = (1)  4   − 2 
  = 2 − 3

  2   2  
5. Evaluate, without using a calculator, leaving in surd form:
1
3 −1
tan 60° − tan 30°
3= =
3
=
1 + tan 30° tan 60°
1+1
 1 
1+ 
 3
 3
3−
( )
tan 60° − tan 30°
1 + tan 30° tan 60°
2
3 = 1
2
3
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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° =
x
from which,
13.0
x = 13.0 sin 70° = 12.22
(b) sin 22° =
x
from which,
15.0
x = 15.0 sin 22° = 5.619
(c) cos 29° =
x
from which,
17.0
x = 17.0 cos 29° = 14.87
(d) cos 59° =
4.30
x
x=
(e) tan 43° =
x
6.0
from which,
x = 6.0 tan 43° = 5.595
(f) tan 53° =
7.0
x
from which,
x=
from which,
4.30
= 8.349
cos 59°
7.0
= 5.275
tan 53°
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© 2014, John Bird
2. Find the unknown sides and angles in the right-angled triangles shown. The dimensions shown
are in centimetres.
(a) By Pythagoras, AC =
tan C =
and
3.0
from which,
5.0
and
 3.0 
∠C = tan −1 
 = 30.96°
 5.0 
∠A =180° – 90° – 30.96° = 59.04°
(b) By Pythagoras, DE =
sin D =
3.02 + 5.02 = 5.831 cm
4.0
8.0
8.02 − 4.02 = 6.928 cm
from which,
 4.0 
∠D = sin −1 
 = 30°
 8.0 
∠F =180° – 90° – 30° = 60°
(c) ∠J =180° – 90° – 28° = 62°
sin 28° =
HJ
from which,
12.0
HJ = 12.0 sin 28° = 5.634 cm
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By Pythagoras, GH = 12.02 − 5.6342 = 10.60 cm
(d) ∠L =180° – 90° – 27° = 63°
sin 27° =
LM
from which,
15.0
LM = 15.0 sin 27° = 6.810 cm
By Pythagoras, KM = 15.02 − 6.8102 = 13.37 cm
(e) ∠N =180° – 90° – 64° = 26°
tan 64° =
NP
4.0
from which,
By Pythagoras, ON =
NP = 4.0 tan 64° = 8.201 cm
8.2012 + 4.02 = 9.124 cm
(f) ∠S =180° – 90° – 41° = 49°
tan 41° =
RS
5.0
from which,
By Pythagoras, QS =
RS = 5.0 tan 41° = 4.346 cm
4.3462 + 5.02 = 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° =
BC
from which, height of building, BC = 2 tan 73° = 6.54 m
2
4. Determine the length x in the diagram.
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© 2014, John Bird
From triangle ABC in the sketch above,
tan 28° =
BC 5
=
AB x
from which, x =
5
= 9.40 mm
tan 28°
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 =
and
Hence,
sin 28° =
56°
= 28°
2
BE BE
=
6
AB
from which, BE = 6 sin 28° = 2.817 m
height, h = 2 × 2.817 = 5.63 m
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© 2014, John Bird
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° =
AB
from which, height of tower, AB = 105 tan 19° = 36.15 m
105
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° =
30
30
from which, distance to aerial, BC =
= 48 m
tan 32°
BC
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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© 2014, John Bird
Tan 19° 50' =
90.0
90.0
from which, distance of boat to the cliff, BC =
= 249.5 m
tan19°50 '
BC
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
= 298.56 m
tan15°
Tan 23° =
80
BC
from which, BC =
80
= 188.47 m
tan 23°
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' =
h
from which, height h = (tan 18°40´)(50 + BD)
50 + BD
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© 2014, John Bird
= (0.337833)(50 + BD) = 16.89165 + 0.337833(BD)
Tan 26°22' =
h
from which, height, h = (tan 26°22´)(BD)
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 =
16.89165
= 107.01 m
0.157846
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° =
AC
from which, AC = 200 tan 32° = 124.97 m
200
Tan 30° =
BC
from which, BC = 200 tan 30° = 115.47 m
200
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.
658
© 2014, John Bird
Tan 26° =
h
h
from which, DC =
= 2.0503h
tan 26°
DC
Tan 31° =
h + 25
h + 25
from which, DC =
= 1.66428(h + 25) = 1.66428h + 41.607
DC
tan 31°
Equating the DC values gives:
i.e.
2.0503h = 1.66428h + 41.607
2.0503h – 1.66428h = 41.607
i.e.
0.3860h = 41.607
from which,
height of cliff, h =
41.607
= 107.8 m
0.3860
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° =
AB
hence, width of road, AB = 4.2 tan 66° = 9.43 m
4.2
CE CE CE
From triangle DEC, tan 34° = = =
from which, CE = 9.43 tan 34° = 6.36 m
DE AB 9.43
659
© 2014, John Bird
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
Tan 24° =
AB
300 − BC
i.e.
from which, AB = BC tan 20°
from which, AB = (300 – BC) tan 24°
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 =
Tan 20° =
133.57
= 165.06 m
0.8092
AB
from which, height of tower, AB = 165.06 tan 20° = 60 m, to the nearest metre
165.06
660
© 2014, John Bird