Question

A tower subtends an angle of ${30}^{\circ }$ at a point, C, on the same level as its foot, A. At a second point, D, 'h' meters above C, the angle of depression of the foot of the tower is ${60}^{\circ }$. What is the height of the tower?

• A. $\frac{h}{2}m$
• B. $\frac{h}{3}m$
• C. $h\sqrt{3}m$
• D. $\frac{h}{\sqrt{3}}m$

Answer 1

The correct option is B

$\frac{h}{3}m$

Let the height of tower be H

In triangle ABC

$\tan {30}^{\circ }=\frac{AB}{AC}=\frac{H}{AC}$ .................................. 1

In triangle ADC

$\tan {60}^{\circ }=\frac{CD}{AC}=\frac{h}{AC}$.................................2

Divide 1 by 2

We get $\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}=\frac{H}{h}$
$Weknow\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}$
$So\frac{H}{h}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{\sqrt{3}\times \sqrt{3}}$

$H=\frac{h}{3}$