A pole stands vertically inside a triangular park ABC.

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 $${\pi  \over 3}$$. If the radius of the circumcircle of $$\Delta$$ABC is 2, then the height of the pole is equal to :

Accepted Answer

Let PD = h, R = 2

As angle of elevation of top of pole from A, B, C are equal. So D must be circumcentre of $$\Delta$$ABC

$$\tan \left( {{\pi  \over 3}} \right) = {{PD} \over R} = {h \over R}$$

$$h = R\tan \left( {{\pi  \over 3}} \right) = 2\sqrt 3 $$

Solutions · 1