A pole stands vertically inside a triangular park ABC. Let the angel of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\triangle ABC$ is 2, then the height of the pole is equal to :

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Heights and Distances

A pole stands vertically inside a triangular park ABC. Let the angel of elevation of the top of the pole from each corner of the park be

\frac{π}{3}

. If the radius of the circumcircle of

△ A B C

is 2, then the height of the pole is equal to :

(A) $\frac{1}{3}$
(B) $3$
(C) $23$ ✓ Correct
(D) $\frac{2 3}{3}$

Answer: (C)

Solution

Equal angles of elevation from

A

B

and

C

mean the foot of the pole is equidistant from all three corners, so it stands at the circumcentre and that common distance is the circumradius

R = 2

. If

h

is the height of the pole, then from any corner

tan \frac{π}{3} = \frac{h}{R} ⟹ h = 2 tan \frac{π}{3} = 23

The pole is

23

units high.

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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