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

We have been given angle of elevation of the from the foot of a tower as 60°, angle of depression from the top of the tower to the foot of the hill is 30° and the height of the tower. We need to find the height of the hill.

Let the height of the hill be ℎ. We know that for any angle θ, $\tan \theta =\frac{\text{ pependicular }}{\text{ base }}$

In the ∆𝐴𝐵𝐶,

$\begin{array}{r}\tan {60}^{\circ }=\frac{h}{x}\\ \Rightarrow \sqrt{3}=\frac{h}{x}\\ \Rightarrow h=\sqrt{3}x\end{array}$

In ∆𝐵𝐶𝐷,

$\begin{array}{r}\tan {30}^{\circ }=\frac{50}{x}\\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{x}\\ \Rightarrow 50\sqrt{3}=x\end{array}$

Solving (𝑖) and (𝑖𝑖), we get 

𝑥 = 50 $\sqrt{3}$ 𝑚 

ℎ = 150𝑚 

Hence, the height of the hill is 150 𝑚.