$A$ pole stands vertically inside a triangular park $\Delta ABC$. 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 $\Delta ABC$ is $2$,then the height of the pole is equal to:

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

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