Question

The angle of elevation of the top of the hill from the foot of a tower is 60 degrees and the angle of depression from the top of the tower to the foot of the hill is 30 degrees. If the tower is 50m high ; find the height of the hill.

Answer 1

Answer:

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 rt.△ABC,

Cos 60° = x/h

x = h cot 60°........(i)

In rt.△DBC,

cot 30° = x/50

x = 50 cot 30°........(ii)

Equating (i) and (ii)

h cot 60° = 50 cot 30°

h = 50 cot 30°/cot 60°

h = 50 × √3/1√3 = 50 × 3 = 150 m

Therefore the height of the hill is 150 m.

Answer 2

Answer:

Let AB is the Tower of height = h = 50 m.

And,  let the Height of Hill CD = H m.

Distance between The root of the tower and hill = BC 

Now,

In ΔABC

∠C = 30°

    TAN(C) =  AB/BC

⇒ TAN(30) =  50/BC

⇒  1/√3 = 50 /BC

⇒ BC = 50√3 m.

Now,

In ΔBCD,

∠B = 60°

    Tan(B) = CD/BC

⇒ Tan(60) = H/BC

⇒ BC√3 = H

⇒ H = 50√3*√3 = 150 m.