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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.`
8 years ago

Pavan kumar
18 Points
```										(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.

```
8 years ago
Ramesh V
70 Points
```										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
--
regards
Ramesh
```
8 years ago
8 Points
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```
8 years ago
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