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

  1. (1) $\frac{2 \sqrt{3}}{3}$

  2. (2) $2 \sqrt{3}$

  3. (3) $\sqrt{3}$

  4. (4) $\frac{1}{\sqrt{3}}$


Correct Option: , 2

Solution:

Let $\mathrm{PD}=\mathrm{h}, \mathrm{R}=2$

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

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

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

Administrator

Leave a comment

Please enter comment.
Please enter your name.