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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
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