A kite is flying with the string inclined at $30^{o}$ to the horizon. The height of the kite above the ground, when the string is $15 m$ long is

15 m

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30 m

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\frac{15}{2} m

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\frac{15}{3} m

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Solution

The correct option is C $\frac{15}{2} m$
$sin 30^{0} = \frac{A B}{A C} = \frac{h}{15}$

$\Rightarrow \frac{1}{2} = \frac{h}{15}$

$\Rightarrow h = \frac{15}{2} m$

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A kite is flying with the string inclined at 30o to the horizon. The height of the kite above the ground, when the string is 15 m long is

The correct option is C 152 msin300=ABAC=h15⇒12=h15⇒h=152 m

A kite is flying with the string inclined at $30^{o}$ to the horizon. The height of the kite above the ground, when the string is $15 m$ long is

The correct option is C $\frac{15}{2} m$
$sin 30^{0} = \frac{A B}{A C} = \frac{h}{15}$

$\Rightarrow \frac{1}{2} = \frac{h}{15}$

$\Rightarrow h = \frac{15}{2} m$