Question

From the foot of a hill the angle of elevation of the tip of a tower is found to be 45°. After walking 2 km upwards along the slope of the hill ,which is inclined at 30°, the same is formed to be 60°. Find the height of the hill along with the tower

Answer 1

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.

Answer 2

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 AABC

ZC = 30°

TAN(C) = AB/BC

⇒ TAN(30) = 50/BC

⇒1/√3 = 50/BC ⇒ BC= 50-√3 m.

Now, In ABCD,

ZB = 60°

Tan(B) = CD/BC

⇒ Tan(60) = H/BC

⇒ BC√3 = H

⇒ H=50-√3*√3 = 150 m.