Question

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill? 

(a) 100 m 

(b) 125 m 

(c) 50√3 m 

(d) 150 m

Answer 1

(d) 150 meters

Let the tower be AB = 50 m and hill be CD of height h meters. 

Given, ∠DBC = 60°, ∠ACB = 30° In ΔABC,

\(\frac{AB}{BC} = \) tan 30° ⇒ \(\frac{50}{BC} = \frac{1}{\sqrt3}\)

⇒ BC = 50√3 metres. 

In ΔDCB,

\(\frac{DC}{CB} = \) tan 60° ⇒ \(\frac{h}{50\sqrt3} = \sqrt3\)

⇒ = 50√3 x √3 = 150 meters.