Question

The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively ${30}^{\circ }$ and ${45}^{\circ }$ . Find the height of the rock.

Answer 1

Let AB be the height of Rock which is H m. and makes an angle of elevations 45° and 30° respectively from the bottom and top of tower whose height is 100m.

Let AE + h, BC = x, CD = 100, $\angle ACB=45,\angle ADE=30$

We have to find the height of the rock

We have the corresponding figure as

So we use trigonometric ratios.

In $△ABC$ ,

$tan45=\frac{AB}{BC}$

$1=\frac{100+h}{x}$

$x=100+h$,

Again in $△ADE$

$tan30=\frac{AE}{DE}$

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

$100+h=\sqrt{3}h$

$h=136.65$

$H=100+136.65=236.65$

Hence the height of rock is 236.65 m