A boy is standing on the ground and is flying a kite which is attached to a 150 m long string at an angle of elevation of 30°. Another boy is standing on the roof of a 25 m high building and is flying his kite at an elevation of 45°. 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.
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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
Hope This Helps :)
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
Hope This Helps :)
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