Introduction to trigonometry 107

Hence height of pylonAB

=80tan23◦= 80 ( 0. 4245 )= 33 .96m

=34m to the nearest metre.

80 m

23 

A

C B

Figure 11.21

Problem 25. A surveyor measures the angle of

elevation of the top of a perpendicular building as

19 ◦. He moves 120m nearer the building and finds

the angle of elevation is now 47◦. Determine the

height of the building.

The buildingPQand the angles of elevation are shown
in Fig. 11.22.

In trianglePQS,

tan19◦=

h

x+ 120

hence h=tan19◦(x+ 120 ),

i.e. h= 0. 3443 (x+ 120 ) (1)

P

Q

h

x

R

S

120

(^478198)
Figure 11.22
In trianglePQR,tan47◦=
h
x
hence h=tan47◦(x),i.e.h= 1. 0724 x (2)
Equating equations (1) and (2) gives:
0. 3443 (x+ 120 )= 1. 0724 x
0. 3443 x+( 0. 3443 )( 120 )= 1. 0724 x
( 0. 3443 )( 120 )=( 1. 0724 − 0. 3443 )x
41. 316 = 0. 7281 x
x=
41. 316
0. 7281
= 56 .74m
From equation (2),height of building,
h= 1. 0724 x= 1. 0724 ( 56. 74 )=60.85m.
Problem 26. The angle of depression of a ship
viewed at a particular instant from the top of a 75m
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
1minute 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 cliffAB, the initial position
of the ship atCand the final position atD. Since the
angle of depression is initially 30◦then∠AC B= 30 ◦
(alternate angles between parallel lines).
tan30◦=
AB
BC

75
BC
hence BC=
75
tan30◦

75
0. 5774
=129.9m
=initial position of ship from
base of cliff
x
75m
308
208
308
208
A
B C D
Figure 11.23
In triangleABD,
tan20◦=
AB
BD

75
BC+CD

75
129. 9 +x
Hence^129.^9 +x=
75
tan20◦

75
0. 3640
= 206 .0m
from which x= 206. 0 − 129. 9 = 76 .1m
Thus the ship sails 76.1m in 1minute, i.e. 60s, hence
speed of ship

distance
time

76. 
    

    1
    60
    m/s

    
    


77. 1 × 60 × 60
    60 × 1000
    km/h=4.57km/h