Question

A tower stands vertically on the ground. From a point on the ground, which is 24 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be ${60}^{\circ }$. Find the height of the tower.

• A. $26\sqrt{3}m$
• B. $24\sqrt{3}m$
• C. $24m$
• D. $26m$

Answer 1

The correct option is B

$24\sqrt{3}m$

The above $\Delta ABC$ represents the given situation.

AB = 24 m

In Triangle ABC,
$tan{60}^{\circ }=\sqrt{3}=\frac{CB}{AB}\Rightarrow AB\times \sqrt{3}=CB\Rightarrow CB=24\sqrt{3}m.$

Hence, the height of the tower = $24\sqrt{3}m.$