Question

A boy is standing on the ground and flying a kite with 75 m of string at an elevation of 45${}^o$. Another boy is standing on the roof of a 25 m high building and is flying his kite at an elevation of 30${}^o$. Both the boys are on opposite sides of the two 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 length of string of first boy's kite and $AD$ be the length of string of second boy's kite.

In $△ADC$

$\tan {30}^o=\frac{AC}{CD}$

=>$\frac{AC}{CD}=\frac{1}{\sqrt{3}}$

=>Let $AC=x,CD=x\sqrt{3}$

$AE=AC+CE=x+25$

In $△AEB$

$\sin {45}^o=\frac{1}{\sqrt{2}}=\frac{AE}{AB}$

=>$\frac{x+25}{75}=\frac{1}{\sqrt{2}}$

=> $x=\frac{75\sqrt{2}-50}{2}=28.03m$

=> $AC=28.03m,CD=48.54m$

$AD=\sqrt{A{C}^2+C{D}^2}=\sqrt{{28.03}^2+{48.54}^2}=\sqrt{785.6809+2356.1316}=\sqrt{3141.8125}=56.05m$