Question

The angle of elevation of the top of a hill at 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, what is the height of the hill?

Answer 1

Let h' be the height of hill AB. And CD be the tower of height 50m. Angle of elevation of the top of hill from the foot of tower is 60° and angle of elevation of top of tower from foot of hill is 30°. Let AB = h, and $\angle DAC=30,\angle ACB=60$

Here we have to find height of hill.

The corresponding figure is as follows

So we use trigonometric ratios.

In $△ACD$

$tan30=\frac{CD}{AC}$

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

$x=50\sqrt{3}$

Again in $△ABC$

$tan60=\frac{AB}{AC}$

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

$h=x\sqrt{3}=150$

Hence the height of hill is 150 m.