A pole stands vertically inside a triangular park Δ ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then in Δ ABC the foot of the pole is at the :
(A) centroid (B) circumcentre
(C) incentre (D) orthocenter.

(D) orthocentre

OD is the vertical pole.

Hence <DOA = <DOB = 90

 It is given that <OAD = <OBD

OD is common to both triangle AOD and BOD

Hence triangles AOD and BOD are congruent.

Therefore OA = OB.

Similarly, OA = OC.

Therefore o is equidistant from A,B and C, and hence is at the orthocentre.

since the angle of elevation of pole is same, which means for a particular height of pole, the distance of pole feet from and any point A,B,C is same

This means that the feet of pole shld be at centroid of triangle ABC

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regards

Ramesh