A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 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 that the second boy must have so that the two kites meet.
Answers
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Given : A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites.
Solution :
In the figure below ‘C’ is the position of the both the kites. Let the length of the second kite string is CD = h and t CF = x m .BF = ED = 10 m .
Now,
In ∆ABC
Sin 30° = BC/AC
½ = (BF + CF)/100
½ = (10 + x)/100
100 = 2(10 + x)
20 + 2x = 100
2x = 100 - 20
2x = 80
x = 80/2
x = 40m ………….(1)
In ∆CFD
sin 45° = CF/CD
1/√2 = x/h
h = √2 x …………….(2)
On substituting value of x = 40 m in eq (2) :
h = √2 × 40
h = 40√2 m
Hence, the length of the string that the second boy must have so that the two kites meet is 40√2 m.
HOPE THIS ANSWER WILL HELP YOU……
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