Question

A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be π/3. If the radius of the circumcircle of ΔABC is 2, then the height of the pole is equal to :

• A. 2$\sqrt{3}$ / 3
• B. 2$\sqrt{3}$
• C. 1 / $\sqrt{3}$
• D. $\sqrt{3}$

Hint:

If the angle of elevation is same from all vertices, the foot of the pole is the circumcenter. If same from all sides, it is the incenter.

Answer 1

The Correct Option is B

Step 1: Since the angle of elevation of the top of the pole ($h$) is the same from all vertices, the foot of the pole must be the circumcenter ($O$) of $\Delta ABC$.
Step 2: In $\Delta OAV$ (where $V$ is the top), $OA = R = 2$ and $\angle OAV = \pi/3$.
Step 3: $\tan(\pi/3) = \frac{h}{OA} \implies \sqrt{3} = \frac{h}{2} \implies h = 2\sqrt{3}$.