Question

A boy is standing on the ground and flying a kite with $150m$ of string at an elevation of ${30}^0$. Another boy is standing on a roof of a $25m$ high building and is flying his kite at an elevation ${45}^{\circ }$. Both the boys are on opposite sides of both the kites. Find the length of the string (in meters), that the second boy must have so that the two kites meet.

Answer 1

Let $C$ be the position of the kite and $A,B$ be the position of first and second boy respectively.

Let the length of the string (CD) with second boy be $xmeters.$

In $△ABC,\sin {30}^o=\frac{BC}{AC}$

$\Rightarrow$ $\frac{1}{2}=\frac{BC}{100}$

$\therefore$ $BC=\frac{100}{2}=50m$

Now, $CF=BC-BF=50m-10m=40m$

In $△CFD,\sin {45}^o=\frac{CF}{CD}$

$\Rightarrow$ $\frac{1}{\sqrt{2}}=\frac{40}{x}$

$\therefore$ $x=40\sqrt{2}m=40\times 1.414=56.56m$ $(Take\sqrt{2}=1.414)$

Hence, the length of the string second by must-have is $56.56m$.