Question

The angle of elevation of the top of a hill from the foot of a tower is ${60}^o$ and the angle of depression from the top of the tower to the foot of the hill is ${30}^o$. 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 ${60}^o$, i.e., $\angle ADB={60}^o$ and the angle of depression of the foot of hill from the top of the tower is ${30}^o$, i.e., $\angle CBD={30}^o$.

Now in right angled $\Delta CBD$:

$tan{30}^o=\frac{CD}{BD}$

$BD=\frac{CD}{tan{30}^o}$

$BD=\frac{50}{\frac{1}{\sqrt{3}}}$

$BD=50\sqrt{3}$ m

In right $\Delta ABD$:

$tan{60}^o=\frac{AB}{BD}$

$AB=BD\times tan{60}^o$

$AB=50\sqrt{3}\times \sqrt{3}$

$AB=50\times 3$

$AB=150$ m

Hence, the height of the hill is $150$ m.