Question

From the top of a building $60$m high the angle of depression of the top and bottom of a tower are observed to be ${30}^{\circ }$ and ${60}^{\circ }$.Find the height of the tower.

Answer 1

Let $AD=h$ be the height of the tower.

$BC$ be the height of the building.

Here $BC=60$m

From the figure,$\angle DCF={30}^{\circ }\Rightarrow \angle EDC={30}^{\circ }$

and $\angle ACF={60}^{\circ }\Rightarrow \angle BAC={60}^{\circ }$

In $△DEC,\tan {30}^{\circ }=\frac{EC}{DE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{DE}$

$\Rightarrow DE=(60-h)\sqrt{3}$

In right angled triangle,$△ABC,\tan {60}^{\circ }=\frac{BC}{AB}=\frac{60}{DE}$

$\Rightarrow \sqrt{3}=\frac{60}{(60-h)\sqrt{3}}$

$\Rightarrow 3(60-h)=60$

$\Rightarrow 180-3h=60$

$\Rightarrow 180-60=3h$

$\Rightarrow 3h=120$

$\Rightarrow h=\frac{120}{3}=40$

$\therefore$ height of the tower is $40$m