Question

A vertical pole stand at a point O on a horizontal ground. A and B are point on the ground 10m apart. The pole subtends an angle 30${}^{\circ }$ and 60${}^{\circ }$ at A and B, respectively, and AB subtends an angle 45${}^{\circ }$ at O. Then height of tower equals

• A. $10\sqrt{3}$
• B. $\frac{5}{2}\sqrt{3}$
• C. $5\sqrt{3}$
• D. None of these

Answer 1

The correct option is C $5\sqrt{3}$

Given that

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

Let $PO$ be the height of the tower

In $△PAO,\tan \angle PAO=\frac{PO}{OA}$

$⟹PO=OA\tan {30}^{\circ }⟹OA=PO\sqrt{3}$

In $△PBO,\tan \angle PBO=\frac{PO}{OB}$

$⟹PO=OB\tan {60}^{\circ }⟹OB=\frac{PO}{\sqrt{3}}$

$AB=10⟹OA-OB=10⟹PO(\sqrt{3}-\frac{1}{\sqrt{3}})=10⟹PO=5\sqrt{3}$

So height of the tower is $5\sqrt{3}m$