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∘ of at the foot of the tower and the angle of elevation of the top of the tower from A or B is 30∘. 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

Answer: (C)

$\frac {a}{\sqrt 3}$

Answer 2

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∘ 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, ∠AOB=60∘ and ∠OAB=∠OBA
∴ΔOAB is an equilateral triangle.
∴OA=OB=AB=a
Now in △OAC,
tan⁡30∘=ha
⇒13=ha [Using trigonometric ratios ]
⇒h=a3
∴ The height of the tower is a3
Hence, the correct option is (C): $\frac {a}{\sqrt 3}$.