Question

The angle of elevation of the top of a hill at the foot of a tower is ${60}^{\circ }$ and the angle of depression from the top of the tower to the foot of the hill is ${30}^{\circ }$. If the tower is 50 m high, find the height of the hill.

Answer 1

Let AB be the hill and CD be the tower.
Angle of elevation of the hill at the foot of the tower is ${60}^{\circ }.i.e.,\angle ADB={60}^{\circ }$ and the angle of depression of the foot of hill from the top of the tower is ${30}^{\circ }$, i.e., $\angle CBD={30}^{\circ }$.
Now in right angled $\Delta CBD$:
tan ${30}^{\circ }=\frac{CD}{BD}\Rightarrow BD=\frac{CD}{tan{30}^{\circ }}\Rightarrow BD=\frac{50}{\left[\frac{1}{\sqrt{3}}\right]}\Rightarrow BD=50\sqrt{3}m$
In right $\Delta ABD$:
tan ${60}^{\circ }=\frac{AB}{CD}$
$\Rightarrow AD=BD\times tan{60}^{\circ }=50\sqrt{3}\times \sqrt{3}m=50\times 3m$
= 150 m
Hence, the height of the hill is 150 m.