Question

A vertical pole stand at a point $O$ on a horizontal ground. $A$ and $B$ are points on the ground $10m$ apart The pole subtends an angle ${30}^{\circ }$ and ${60}^{\circ }$ at A and B respectively . Then height of tower equals

• A. $10\sqrt{3}$ m
• B. $5\sqrt{3}$ m
• C. $15\sqrt{3}$ m
• D. None of these

Answer 1

The correct option is B $5\sqrt{3}$ m

Given that:

$\angle PAO={30}^{\circ },\angle PBO={60}^{\circ },AB=10m$

To find:

Height of the tower $h=?$

Solution:

Let $PO$ be the height of the tower.

In $△PAO$

$\tan \angle PAO=\frac{PO}{OA}$

or, $\tan {30}^{\circ }=\frac{h}{OA}$

or, $OA=\sqrt{3}h$ $............\left(i\right)$

In $△PAO$

$\tan \angle PBO=\frac{PO}{OB}$

or, $\tan {60}^{\circ }=\frac{h}{OB}$

or, $OB=\frac{h}{\sqrt{3}}$ $............\left(ii\right)$

From eqn. (i) and (ii),

$AB=OA-OB$

or, $10m=\sqrt{3}h-\frac{h}{\sqrt{3}}$

or, $10\sqrt{3}m=3h-h$

or, $2h=10\sqrt{3}m$

or, $h=5\sqrt{3}m$

So, Height of the tower $=5\sqrt{3}$

Hence, $B$ is the correct option.