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

Let $AD$ be the height of tower and $BC$ be the height of the hill.

Given $\angle CAB=60^{\circ },\angle ABD=30^{\circ }$ and $AD=50$m.

Let $BC=h$ metres.

Now, in the right angled $△DAB,\tan 30^{\circ }=\frac{AD}{AB}$

$\Rightarrow AB=\frac{AD}{\tan 30^{\circ }}$

$\therefore AB=50\sqrt{3}$m                (1)

Also, in the right angled $△CAB,\tan 60^{\circ }=\frac{BC}{AB}$

$\Rightarrow BC=AB\tan 60^{\circ }$

Thus, using (1) we get $h=BC=(50\sqrt{3})\sqrt{3}=150$m

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