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


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Given:

Height of the tower, h = DC = 50 m

Angle of elevation of a hill at the foot of tower, θ = 60°

Angle of elevation of top of tower from foot of the hill, ϕ = 30 °

To find: Height of the hill

Solution:

Let h be the height of the hill and x m be the distance between the foot of the hill and foot of the tower.


In right angled Δ ABC,


cos 60 o = x / h


x = h cot 60 o …………… (i)


In right angled Δ DBC,


cot 30 o =
x / 50


x = 50 cot 30 o …………… (ii)


Equating (i) and (ii)


h cot 60° = 50 cot 30°


h = 50 cot 30° / cot 60°

= 50 × 3

= 150 m


Therefore the height of the hill is 150 m.

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