A pole stands vertically, inside a triangular
park $ABC$. If the angle of elevation of the top of the pole
from each corner of the park is same, then in $∆ABC$, the foot
of the pole is at the

Let$OP$ be the pole with $O$ as the foot, and
the angle of elevation of the top $P$of the pole at each of the
corner $A, B, C$ be a. (Fig. 27.14)

then                  $\frac{OP}{OA}=\tan \alpha$

                       $\Rightarrow OA=OP\cot \alpha$

Similarly          $OB=OC=OP\cot \alpha$

Since              $OA=OB=OC,O$   is the circumcentre of the triangle

$ABC.$