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Question
A boy is standing on the ground and flying a kite with 100m of sting at an elevation of 30°. Another boy is standing on the roof of a 10m 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.
Solution
Let C be the position of the first boy and D be the position of the second boy who is standing on the roof of a 10 m high building.
Let B be the position of the kites of both the boys.
Let AB = h and CA = x.
In ΔABC,
sin30° = `"h"/100`
⇒ `1/2 = "h"/100`
⇒ h = 50 ... (1)
In ΔBDE,
`tan45^circ = "BE"/"BD"`
⇒ `1 = ("h" - 10)/"X"`
⇒ X = (h - 10) ...(2)
From (1) and (2),
x = 50 - 10 = 40
In ΔBDE,
`sin45^circ = "BE"/"BD"`
⇒ `1/sqrt(2) = ("h" - 10)/("BC")`
⇒ `"BC" = sqrt(2)(50-10) = 40sqrt(2)`
Thus, the required length of the string that the second boy must have `40sqrt(2)` m
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