Question

A tower stands at the center of a circular park. $A$ and $B$ are two points on the boundary of the park such that $AB=a$ subtends an angle ${60}^{\circ }$ of at the foot of the tower and the angle of elevation of the top of the tower from $A$ or $B$ is ${30}^{\circ }$. The height of the tower is

• A. \(\frac{2a}{ \sqrt{3} }\)
• B. \(2a\sqrt 3\)
• C. \(\frac {a}{\sqrt 3}\)
• D. \(a\sqrt 3\)

Answer 1

The Correct Option is C

Explanation:
Let $OC$ be the tower at the centre $O$ of the circular park.
$A$ and $B$ are two points on the boundary of the park ($=a$) subtends an angle of ${60}^{\circ }$ at the foot of the tower and the angle of elevation of the top of the tower from $A$ or $B$ is
Let '$h$' be the height of the tower.
We have to find the value of '$h$'.

Since, $\mathrm{\angle }AOB={60}^{\circ }$ and $\mathrm{\angle }OAB=\mathrm{\angle }OBA$
$\therefore \mathrm{\Delta }OAB$ is an equilateral triangle.
$\therefore OA=OB=AB=a$
Now in $\mathrm{△}\mathrm{OAC}$,
$\tan {30}^{\circ }=\frac{h}{a}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{a}$ [Using trigonometric ratios ]
$\Rightarrow h=\frac{a}{\sqrt{3}}$
$\therefore$ The height of the tower is $\frac{a}{\sqrt{3}}$
Hence, the correct option is (C): \(\frac {a}{\sqrt 3}\).