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

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

Answer 1

The correct option is C

$2\surd 3$

Explanation for the correct option:

Find the height of the pole:

Let PD = h, R = 2

As we can see from the figure angle of elevation of top of pole from A, B, C are equal So D must be circumcentre of △ABC

$\begin{array}{rcl}& \Rightarrow & \tan \left(\frac{\mathrm{\pi }}{3}\right)=\frac{PD}{R}\\ & =& \frac{h}{R}\end{array}$

$\Rightarrow$ $h=R\tan \left(\frac{\mathrm{\pi }}{3}\right)$

$=2\sqrt{3}$

$\therefore$The height of the pole is $2\sqrt{3}$

Hence, Option ‘C’ is Correct.