Question

A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is sanding 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.

Answer 1

Let AB be the string of string x m. let DF be the ground and a boy flying the kite of 100 m string at an elevation of 30 °. And another boy flying the kite of 10 m high building at an angle of elevation of 45°.

Let AE = H, AC = h, CE = 10, AB = x and AF = 100 m 

∠ABC = 45°, ∠AFE = 30°

Here we have to find length of string.

We use trigonometric ratios.

`In ΔAFE`

 

`=> sin 30^@ = (AE)/(AF)`

`=> 1/2 = H/100`

=> h = 50

=> h = H - 10

=> h = 50 - 10

=>  h = 40

Again in ΔABC

`=> sin 45^@ = (AB)/(AC)`

`=> 1/sqrt2 = h/x`

`=> 1/sqrt2 = 40/x`

`=> x = 40sqrt2`

Hence the leght of string is `40sqrt2`