Question

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take √3 = 1.732)

Answer 1

In triangle OAC tan 45=AC/OA=1

AC=OA=80

IN TRIANGLE ODB

tan 30=BD/OB=1/√ 3

OB=BD√ 3-AC√ 3(since ac=bd as bird remains constant height)

AB=OB-OA=AC√ 3-AC=AC(√ 3-1)=80(√ 3-1)

NOW FOR 2 SEC

80(√ 3-1)/2=40(√ 3-1)

=29.2M/SEC


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