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?
In the fig DC is the tower and AB is the hill.

In ∆DCB
tan 30° = ![]()
= ![]()
BC = 50√3 m
In ∆ABC
tan 60° = ![]()
√3 = ![]()
AB = 50√3× √3 m ⇒ 150m
Therefore the distance between tower and hill is 50√3m and height of hill is 150m
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