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

Let the height of the pole be

$$h$$

and let the foot of the pole be at point $$P$$ inside triangle $$ABC.$$

Given that the angle of elevation of the top of the pole from each vertex is

$$\frac\pi3$$

From vertex $$A,$$

$$\tan\frac\pi3=\frac{h}{AP}$$

$$\sqrt3=\frac{h}{AP}$$

$$AP=\frac{h}{\sqrt3}$$

Similarly,

$$BP=\frac{h}{\sqrt3}$$

and

$$CP=\frac{h}{\sqrt3}$$

Hence,

$$AP=BP=CP$$

Therefore, point $$P$$ is the circumcenter of triangle $$ABC.$$

Thus,

$$AP=R$$

where $$R$$ is the circumradius.

Given,

$$R=2$$

Hence,

$$\frac{h}{\sqrt3}=2$$

$$h=2\sqrt3$$

Therefore, the height of the pole is

$$\boxed{2\sqrt3}$$