Question

If angle of elevation of top of a tower $AB$ from a point $P$ on the ground $15$ meter away from the foot is ${60}^o$, find the height of the tower?

• A. $\frac{15}{\sqrt{3}}m$
• B. $15m$
• C. $15\sqrt{3}$
• D. $30m$

Answer 1

The correct option is C $15\sqrt{3}$

Let the height of the tower be $hm$

In $△ABP,$

$BP=15m$

$\tan {60}^{\circ }=\frac{h}{15}$

$⟹h=15\sqrt{3}m$

So height of the tower is $15\sqrt{3}m$