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A kite is flying with the string inclined at $75 °$ to the horizon. If the length of the string is $25 m$ , then height of the kite is

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(\frac{25}{2}) {(\sqrt{3} - 1)}^{2}

(\frac{25}{2}) (\frac{\sqrt{3} + 1}{\sqrt{2}})

(\frac{25}{2}) {(\sqrt{3} + 1)}^{2}

(\frac{25}{2}) (\sqrt{6} + \sqrt{2})

answer is B.

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Detailed Solution

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

75 °

to the horizon.
According to the given data the figure has been done below.
Question Image

Let H be the height of the kite.
In

Δ A B C,

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 75 ° = \frac{AB}{AC}

\sin 75 °

= \frac{H}{25}

H = 25 \times \sin 75 °

…..(1)

\Rightarrow \sin 75 ° = \sin (45 ° + 30 °)

\Rightarrow \sin 45 ° \cos 30 ° + \cos 45 ° \sin 30 °

∵ sin (A + B) = sin A cos B + cos A sin B

⇒ 12 × 32 + 12 × 12

∵ sin 45 ° = 12 and cos30 ° = 32 cos 45 ° = 12 and sin30 ° = 12

\Rightarrow \frac{\sqrt{3} + 1}{2 \sqrt{2}}

Put value of

\sin 75 °

in equation (1),

H = 25 [\frac{\sqrt{3} + 1}{2 \sqrt{2}}]

\Rightarrow H = \frac{25}{2} [\frac{\sqrt{3} + 1}{\sqrt{2}}]

So the height of the kite is

\frac{25}{2} [\frac{\sqrt{3} + 1}{\sqrt{2}}]

m.
So the correct option is 2.

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