(B) Let the height of the pole be $h = PD$,where $P$ is the top of the pole and $D$ is the base of the pole on the ground.Since the angle of elevation

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(B) Let the height of the pole be $h = PD$,where $P$ is the top of the pole and $D$ is the base of the pole on the ground.Since the angle of elevation of the top of the pole $P$ from each corner $A, B, C$ is the same $(\frac{\pi}{3})$,the distances from $D$ to each vertex $A, B, C$ must be equal.Thus,$DA = DB = DC = R$,where $R$ is the circumradius of $\Delta ABC$.Given $R = 2$.In the right-angled triangle $\Delta PDA$,we have:$\tan\left(\frac{\pi}{3}\right) = \frac{PD}{DA} = \frac{h}{R}$$h = R \tan\left(\frac{\pi}{3}\right) = 2 \times \sqrt{3} = 2\sqrt{3}$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Medium

$A$ 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:

JEE MAIN 2021 All JEE MAIN

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

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$A$ 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:

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