JEE Math Practice Question
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot–1 (3$$\sqrt 2 $$ ) and cosec–1 (2$$\sqrt 2 $$ ) respectively, then the height of the tower (in metres) is :
- A.$${{100} \over {3\sqrt 3 }}$$
- B.25
- C.20
- D.10$$\sqrt 5 $$
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