Question

The angle of elevation of the top of a hill from the foot of a tower is ${60}^o$ and the angle of elevation of the top of the tower from 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 tower and CD be the hill. Then, $\angle ACB={30}^o,\angle CAD={60}^o$ and $AB=50m$.

Let $CD=xm$

In right $△BAC$, we have,

$\cot {30}^o=\frac{AC}{AB}$

$\sqrt{3}=\frac{AC}{50}$

$AC=50\sqrt{3}m$

In right $△ACD$, we have,

$\tan {60}^o=\frac{CD}{AC}$

$\sqrt{3}=\frac{x}{50\sqrt{3}}$

$x=50\times 3=150m$

Therefore, the height of the hill is 150 m.