Question

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.

Answer 1

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 $h=BC=\left(50\sqrt{3}\right)\sqrt{3}=150$m

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