Question

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

Answer 1

Let the height of the hill, $CD$ be $hm$.

Height of the tower, $AB=50m$.

In $△BCD,\tan {60}^o=\frac{CD}{BC}$

$\Rightarrow$ $\sqrt{3}=\frac{h}{BC}$

$\Rightarrow$ $BC=\frac{h}{\sqrt{3}}$ -------- ( 1 )

In $△ABC,\tan {30}^o=\frac{AB}{BC}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{50}{BC}$

$\therefore$ $BC=50\sqrt{3}$ -------- ( 2 )

Comparing ( 1 ) and ( 2 ), we get

$\Rightarrow$ $\frac{h}{\sqrt{3}}=50\sqrt{3}$

$\Rightarrow$ $h=50\sqrt{3}\times \sqrt{3}$

$\therefore$ $h=150m$

Thus, the height of the hill is $150m$