a boy isstanding on the ground and flying a kite with a string of length 150m at an angle of elevation of 30 degree. another boyis standing on the roof of a 25 m high building and is flying his kte at an elevation of 45 degree.both the boys are on opposite sides of both the kites find the length of the string( in metres) correct to two decimal places that the second boy must have so that the two kites meet

Asked by ajayrath7 | 13th Aug, 2019, 06:42: PM

Expert Answer:

According to the given information in the question AB = 150 m, 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
BF = BE – EF       
= 75 – 25       
= 50 m

Consider, right angled ΔBFC 
sin 45° = (BF/BC) (1/√2) 
           = (50/BC)
Hence, BC = 50√2
Hence, the length of string required by the second boy such that the both kites meet is 50√2 m.

Answered by Sneha shidid | 14th Aug, 2019, 09:58: AM

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