Question
A boy is standing on the ground and flying a kite with a string of length 150 m, 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 mts), correct to two decimal places, that the second boy must have so that the two kites meet.
Solution
Let A be the boy and F is the position of kite. Let BF be vertical height of the kite and AF is the length of the string i.e. AF = 150 m. It is given that ∠BAF = 30°.
In right triangle ABF, we have
In right triangle ABF, we have
Let D be the position of second boy and DF be the length of second kite. It is given ∠EDF = 45°.
In right triangle DEF, we have
Hence, the length of the string = 
