A kite is flying with the string inclined at $45^{\circ}$ to the horizontal If the string is straight and 50 m long the height at which the kite is flying is

25 \sqrt{2} m

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50 \sqrt{2} m

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25 m

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50 m

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Solution

The correct option is A $25 \sqrt{2} m$
$sin 45^{\circ} = \frac{h}{50}$ or $\frac{1}{\sqrt{2}} = \frac{h}{50}$
or h= $\frac{50}{\sqrt{2}} = \frac{2 \times 25}{\sqrt{2}} = 25 \sqrt{2}$

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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

The correct option is A 25√2msin45∘=h50 or 1√2=h50or h=50√2=2×25√2=25√2

A kite is flying with the string inclined at $45^{\circ}$ to the horizontal If the string is straight and 50 m long the height at which the kite is flying is

The correct option is A $25 \sqrt{2} m$
$sin 45^{\circ} = \frac{h}{50}$ or $\frac{1}{\sqrt{2}} = \frac{h}{50}$
or h= $\frac{50}{\sqrt{2}} = \frac{2 \times 25}{\sqrt{2}} = 25 \sqrt{2}$