A boy is standing on the ground and flying a kite with $ 100 \mathrm{~m} $ of string at an elevation of $ 30^{\circ} $. Another boy is standing on the roof of a $ 10 \mathrm{~m} $ high building and is flying his kite at an elevation of $ 45^{\circ} $. 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.

Given:

A boy is standing on the ground and flying a kite with \( 100 \mathrm{~m} \) of string at an elevation of \( 30^{\circ} \).

Another boy is standing on the roof of a \( 10 \mathrm{~m} \) high building and is flying his kite at an elevation of \( 45^{\circ} \).

Both the boys are on opposite sides of both the kites. 

To do:

We have to find the length of the string that the second boy must have so that the two kites meet.

Solution:

Let $F$ be the kite, $A$ and $C$ be the two boys flying kites.

Boy $C$ is standing on a on the roof of a \( 10 \mathrm{~m} \) high building.

The string $AF$ of kite of boy A is $100\ m$.

Let $h$ be the height of the kite from the ground and $x$ be the length of string of kite of second boy $C$.

$\mathrm{FE}=(h-10) \mathrm{m}$

In $\triangle \mathrm{AFD}$,

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{\mathrm{DF}}{\mathrm{AF}}$

$\sin 30^{\circ}=\frac{h}{100}$

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

$\Rightarrow h=\frac{100}{2}$

$\Rightarrow h=50 \mathrm{~m}$

Similarly,

In $\Delta \mathrm{FEC}$,
$\sin 45^{\circ}=\frac{\mathrm{FE}}{\mathrm{FC}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{50-10}{x}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{40}{x}$

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

$\Rightarrow x=40(1.414)$

$\Rightarrow x=45.656 \mathrm{~m}$

Therefore, the length of the string that the second boy must have so that the two kites meet is $45.656\ m$.

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Updated on: 10-Oct-2022

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