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A kite is flying with the string inclined at $45 °$ to the horizontal. If the string is straight and 50 m long the height at which the kite is flying is

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25 \sqrt{2} m

50 \sqrt{2} m

25 m

50 m

answer is A.

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Detailed Solution

It is given that a kite is flying with the string inclined at

45 °

to the horizontal.
Here h is the height of the kite which is flying with the string inclined at

45 °

to the horizontal.
In the given figure AC represents the string which is 50 m long.
Question Image

ΔABC,

sin θ = P e r p e n d i c u l a r H y p o t e n u s e

\sin 45 ° = \frac{AB}{AC}

\Rightarrow

\sin 45 ° = \frac{h}{50}

\Rightarrow \frac{1}{\sqrt{2}} = \frac{h}{50}

∵ sin 45 ° = 12

\Rightarrow h = \frac{50}{\sqrt{2}}

\Rightarrow h = \frac{2 \times 25}{\sqrt{2}}

\Rightarrow h = 25 \sqrt{2}

So we got the value of h is

25 \sqrt{2}

m.
So the height at which the kite is flying is

25 \sqrt{2}

m.
So the correct option is 1.

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