Animated Solution for Mathematics - Trigonometry: A pole stands vertically inside a triangular park $\Delta ABC$. If the angle of elevation of the top of the pole from each corner of the park is same, then in $\Delta ABC$ the foot of the pole is at the

Imagine a triangular park $\Delta ABC$ situated on a horizontal plane. A vertical pole of height $h$ stands at a point $O$ on the ground, with its top at point $P$.

Since the pole is vertical, it is perpendicular to the ground. This implies that the line segment $OP$ is perpendicular to any line in the plane passing through $O$.

Consequently, we form three right-angled triangles: $\Delta POA$, $\Delta POB$, and $\Delta POC$. In each of these triangles, the angle at $O$ is $90^{\circ }$, and the height $OP$ is common to all.

The problem states that the angle of elevation from each corner ($A$, $B$, and $C$) to the top of the pole $P$ is identical. Let this constant angle be denoted by $\alpha$.

In the right-angled triangle $\Delta POA$, we apply the definition of the tangent function:

Rearranging this expression to solve for the distance from the foot of the pole to the vertex $A$, we obtain:

Since the height $OP$ and the angle $\alpha$ are identical for all three vertices, the same logic applies to triangles $\Delta POB$ and $\Delta POC$. Therefore, we find:

We have arrived at the conclusion that the foot of the pole $O$ is equidistant from all three vertices of the triangular park.

In Euclidean geometry, the unique point that is equidistant from all three vertices of a triangle is defined as the circumcentre of the triangle.

Conclusion: The foot of the pole $O$ must be the circumcentre of $\Delta ABC$.

This result is a powerful tool in coordinate geometry and 3D visualization. Whenever a point in space subtends the same angle of elevation to all vertices of a polygon, the projection of that point onto the plane is the circumcentre of the polygon.