Question

A boy is standing on the ground and flying kite with a string of 150 m,at an angle of elavation 300. Another boy standing on a 25 m high building and his flying his kite at an elavation 450. Both the boys are on opposite sides of both the kites . Find the length of the string in metres that the second boy 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

This is the answer to the question......

Hope this helps you mate......