Question

A boy is standing on the ground and flying kite with a string of 150 M at an angle of elevation of 30°.another boy standing on the roof of a 25 m high building and is playing his right at an angle of elevation of 45 degree. both the boys are on opposite sides of both the kites. find the length of the string that the second one must have so that the two kites meet.

Answer 1

Given length of the string of first kite, AB = 150 m Height of the building where the second boy is flying the kite, CD = EF = 25 m BC is the length of the string of second kite such that both the kites meet at B. In right angled ΔAEB, sin 30° = (BE/150) BE = 150 x (1/2) = 75 m From the figure, BF = BE – EF       = 75 – 25       = 50 m Consider, right angled ΔBFC  sin 45° = (BF/BC) (1/√2)  = (50/BC) Therefore BC = 50√2 Thus the length of string required by the second boy such that the both kites meet is 50√2 m  

Answer 2

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