JEE Main | 2014Trigonometric EquationHard
Question
A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45o. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30o. Then the speed (in m/s) of the bird is
Options
A.20√2
B.20(√3 - 1)
C.40(√2 - 1)
D.40(√3 - √2)
Solution
t = 1 s
From figure tan 45o =
and tan 30 o =
so, y = 20(√3 - 1)
i.e., speed = 20(√3 - 1) m/s.
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