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 :
Correct Option: , 2
Solution:
Let $P D=h, R=2$
As angle of elevation of top of pole from
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are equal So
D must be circumcentre
of $\triangle \mathrm{ABC}$
$\tan \left(\frac{\pi}{3}\right)=\frac{\mathrm{PD}}{\mathrm{R}}=\frac{\mathrm{h}}{\mathrm{R}}$
$h=R \tan \left(\frac{\pi}{3}\right)=2 \sqrt{3}$
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