Introduction to trigonometry 107
Hence height of pylon AB
= 80 tan 23
◦
= 80(0.4245) = 33.96 m
= 34 m to the nearest metre.
80 m
23Њ
A
B
C
Figure 11.21
Problem 25. A surveyor measures the angle of
elevation of the top of a perpendicular building as
19 ◦ . He moves 120 m nearer the building and finds
the angle of elevation is now 47 ◦ . Determine the
height of the building.
The building PQ and the angles of elevation are shown
in Fig. 11.22.
In triangle PQS,
tan 19
◦
=
h
x + 120
hence
h = tan 19
◦
(x + 120),
i.e.
h = 0.3443(x + 120)
(1)
P
Q
h
x
R
S
120
478
198
Figure 11.22
In triangle PQR, tan47 ◦ =
h
x
hence
h = tan 47
◦
(x), i.e. h = 1.0724x
(2)
Equating equations (1) and (2) gives:
0.3443(x + 120) = 1.0724x
0.3443x + (0.3443)(120) = 1.0724x
(0.3443)(120) = (1.0724 − 0.3443)x
41.316 = 0.7281x
x =
41.316
0.7281
= 56.74 m
From equation (2), height of building,
h = 1.0724x = 1.0724(56.74) = 60.85 m.
Problem 26. The angle of depression of a ship
viewed at a particular instant from the top of a 75 m
vertical cliff is 30 ◦ . Find the distance of the ship
from the base of the cliff at this instant. The ship is
sailing away from the cliff at constant speed and
1 minute later its angle of depression from the top of
the cliff is 20 ◦ . Determine the speed of the ship
in km/h.
Figure 11.23 shows the cliff AB, the initial position
of the ship at C and the final position at D. Since the
angle of depression is initially 30 ◦ then ∠AC B = 30 ◦
(alternate angles between parallel lines).
tan 30
◦
=
AB
BC
=
75
BC
hence
BC =
75
tan 30 ◦ =
75
0.5774
= 129.9 m
= initial position of ship from
base of cliff
x
75 m
308
208
308
208
A
B
D
C
Figure 11.23
In triangle ABD,
tan 20
◦
=
AB
BD
=
75
BC + CD
=
75
129.9 + x
Hence 129.9 + x =
75
tan 20 ◦ =
75
0.3640
= 206.0 m
from which
x = 206.0 − 129.9 = 76.1 m
Thus the ship sails 76.1 m in 1 minute, i.e. 60 s, hence
speed of ship
=
distance
time
=
76.1
60
m/s
=
76.1 ×60 × 60
60 ×1000
km/h = 4.57 km/h
Hence height of pylon AB
= 80 tan 23
◦
= 80(0.4245) = 33.96 m
= 34 m to the nearest metre.
80 m
23Њ
A
B
C
Figure 11.21
Problem 25. A surveyor measures the angle of
elevation of the top of a perpendicular building as
19 ◦ . He moves 120 m nearer the building and finds
the angle of elevation is now 47 ◦ . Determine the
height of the building.
The building PQ and the angles of elevation are shown
in Fig. 11.22.
In triangle PQS,
tan 19
◦
=
h
x + 120
hence
h = tan 19
◦
(x + 120),
i.e.
h = 0.3443(x + 120)
(1)
P
Q
h
x
R
S
120
478
198
Figure 11.22
In triangle PQR, tan47 ◦ =
h
x
hence
h = tan 47
◦
(x), i.e. h = 1.0724x
(2)
Equating equations (1) and (2) gives:
0.3443(x + 120) = 1.0724x
0.3443x + (0.3443)(120) = 1.0724x
(0.3443)(120) = (1.0724 − 0.3443)x
41.316 = 0.7281x
x =
41.316
0.7281
= 56.74 m
From equation (2), height of building,
h = 1.0724x = 1.0724(56.74) = 60.85 m.
Problem 26. The angle of depression of a ship
viewed at a particular instant from the top of a 75 m
vertical cliff is 30 ◦ . Find the distance of the ship
from the base of the cliff at this instant. The ship is
sailing away from the cliff at constant speed and
1 minute later its angle of depression from the top of
the cliff is 20 ◦ . Determine the speed of the ship
in km/h.
Figure 11.23 shows the cliff AB, the initial position
of the ship at C and the final position at D. Since the
angle of depression is initially 30 ◦ then ∠AC B = 30 ◦
(alternate angles between parallel lines).
tan 30
◦
=
AB
BC
=
75
BC
hence
BC =
75
tan 30 ◦ =
75
0.5774
= 129.9 m
= initial position of ship from
base of cliff
x
75 m
308
208
308
208
A
B
D
C
Figure 11.23
In triangle ABD,
tan 20
◦
=
AB
BD
=
75
BC + CD
=
75
129.9 + x
Hence 129.9 + x =
75
tan 20 ◦ =
75
0.3640
= 206.0 m
from which
x = 206.0 − 129.9 = 76.1 m
Thus the ship sails 76.1 m in 1 minute, i.e. 60 s, hence
speed of ship
=
distance
time
=
76.1
60
m/s
=
76.1 ×60 × 60
60 ×1000
km/h = 4.57 km/h