Higher Engineering Mathematics

120 GEOMETRY AND TRIGONOMETRY

(Note, ∠PRQ is also φ—alternate angles

between parallel lines.)

Problem 9. An electricity pylon stands on hori-

zontal ground. At a point 80 m from the base of

the pylon, the angle of elevation of the top of the

pylon is 23◦. Calculate the height of the pylon

to the nearest metre.

Figure 12.16 shows the pylonABand the angle of

elevation ofAfrom pointCis 23◦

tan 23◦=

AB

BC

=

AB

80

Hence height of pylonAB

=80 tan 23◦=80(0.4245)= 33 .96 m

=34 m to the nearest metre

Figure 12.16

Problem 10. A surveyor measures the angle of

elevation of the top of a perpendicular build-

ing 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 buildingPQand the angles of elevation are
shown in Fig. 12.17.

In trianglePQS,

tan 19◦=

h

x+ 120

hence h=tan 19◦(x+120),

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

Figure 12.17

In triangle PQR, tan 47◦=

h

x

hence h=tan 47◦(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 .74 m

From equation (2),height of building,

h= 1. 0724 x= 1 .0724(56.74)=60.85 m.

Problem 11. 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 12.18 shows the cliffAB, the initial position

of the ship atCand the final position atD. Since the

angle of depression is initially 30◦then∠ACB= 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

Figure 12.18

In triangleABD,

tan 20◦=

AB

BD

=

75

BC+CD

=

75

129. 9 +x