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 ΔABC, the foot of the pole is at the:
Circumcentre
Understanding the Problem: Pole Elevation in a Triangular Park
The question describes a scenario where a vertical pole is located inside a triangular park, referred to as ΔABC. We are told that the angle of elevation of the top of the pole is the same when viewed from each of the three corners of the park: A, B, and C.
Let's visualize this. Imagine the pole stands vertically with its foot at point P inside the triangle, and its top at point T. The height of the pole is PT. The angle of elevation from a corner, say A, is the angle formed between the line segment PA (on the ground) and the line segment TA (from the corner to the top of the pole). Let this angle be θ.
We are given that the angle of elevation from A, B, and C is the same, i.e.,
- Angle of elevation from A = θ
- Angle of elevation from B = θ
- Angle of elevation from C = θ
Let the height of the pole be \(h\). For corner A, we have a right-angled triangle formed by the pole, the ground distance PA, and the line TA. In this right-angled triangle, the tangent of the angle of elevation is given by:
\(\tan(\theta) = \frac{\text{Height of pole}}{\text{Distance from foot of pole to corner}} = \frac{h}{PA}\)
Similarly, for corners B and C, we have:
\(\tan(\theta) = \frac{h}{PB}\)
\(\tan(\theta) = \frac{h}{PC}\)
Since the angle of elevation \(\theta\) and the height of the pole \(h\) are the same for all three corners, the distances from the foot of the pole (point P) to each corner (A, B, and C) must be equal.
Thus, \(PA = PB = PC\).
Identifying the Foot of the Pole's Location in ΔABC
We have determined that the foot of the pole is a point P inside (or potentially on the boundary of) ΔABC that is equidistant from the three vertices A, B, and C.
Now, let's consider the special points related to a triangle and their distance properties:
- Centroid: This is the intersection point of the medians (lines from a vertex to the midpoint of the opposite side). The centroid is the center of mass but is not generally equidistant from the vertices.
- Circumcentre: This is the intersection point of the perpendicular bisectors of the sides of the triangle. The circumcentre is the center of the circumcircle, which passes through all three vertices of the triangle. By definition, the circumcentre is equidistant from the three vertices. This distance is the radius of the circumcircle.
- Incentre: This is the intersection point of the angle bisectors. The incentre is the center of the incircle, which is tangent to all three sides of the triangle. The incentre is equidistant from the sides of the triangle, not the vertices.
- Orthocentre: This is the intersection point of the altitudes (lines from a vertex perpendicular to the opposite side). The orthocentre has no general property of being equidistant from the vertices or sides.
Comparing the property we found (\(PA = PB = PC\)) with the properties of these special points, it is clear that the foot of the pole must be located at the Circumcentre of the triangle ΔABC, as the circumcentre is the unique point equidistant from the vertices of a triangle.
Analyzing the Options
Based on our analysis, let's look at the provided options:
- Centroid: Incorrect. The centroid is not equidistant from the vertices.
- Circumcentre: Correct. The circumcentre is equidistant from the vertices.
- Incentre: Incorrect. The incentre is equidistant from the sides.
- Orthocentre: Incorrect. The orthocentre is not equidistant from the vertices.
Conclusion
Since the angle of elevation of the top of the pole from each corner of the park ΔABC is the same, the foot of the pole must be equidistant from the vertices A, B, and C. The point in a triangle that is equidistant from its vertices is the circumcentre.
Revision Table: Special Points in a Triangle
| Special Point | Intersection of | Equidistant from | Key Property in this Problem Context |
|---|---|---|---|
| Centroid | Medians | - | Not equidistant from vertices |
| Circumcentre | Perpendicular Bisectors | Vertices | Equidistant from vertices (Matches requirement) |
| Incentre | Angle Bisectors | Sides | Not equidistant from vertices |
| Orthocentre | Altitudes | - | Not equidistant from vertices |
Additional Information: Circumcenter and Triangle Types
The location of the circumcenter depends on the type of triangle:
- Acute Triangle: The circumcenter lies inside the triangle.
- Right Triangle: The circumcenter lies on the midpoint of the hypotenuse.
- Obtuse Triangle: The circumcenter lies outside the triangle.
In this problem, since the pole is stated to be "inside" the triangular park, this implies that the circumcenter must also lie inside the triangle, which means ΔABC must be an acute triangle. However, the core condition derived from the equal angles of elevation is solely about the equidistance from the vertices, which uniquely identifies the circumcenter regardless of whether it's inside or outside the triangle.
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