Question

The angle of elevation of the top of a tower from a point on the ground, which is $30\text{m}$ away from the foot of the tower, is ${30}^{\circ }$. Find the height of the tower.

Answer 1

In the above figure, $PR$ is the tower, and $Q$ is the point of observation at a distance of $30m$ from the base of the tower.

In $△PQR,$

$\tan {30}^{\circ }=\frac{PR}{PQ}$

$\Rightarrow PR=PQ\tan {30}^{\circ }$

$\Rightarrow PR=\frac{30}{\sqrt{3}}$

$\Rightarrow PR=10\sqrt{3}\text{m}$

$\therefore$ Height of the tower is $10\sqrt{3}\text{m}$.