A kite is flying on a string of length 40 $\sqrt{3}$ m. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^{\circ}$ . Find the height at which the kite is flying.

60√3 m

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60 m

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40 m

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20 m

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Solution

The correct option is B
60 m

Length of string AC = 40 $\sqrt{3}$

The inclination of string with the ground is $60^{\circ}$

So, the required triangle is as follows

$s i n 60^{\circ}$ = $\frac{h}{40 \sqrt{3}}$

$\frac{\sqrt{3}}{2}$ = $\frac{h}{40 \sqrt{3}}$

h = 60 m

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A kite is flying on a string of length 40√3m. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60∘. Find the height at which the kite is flying.

The correct option is B 60 m Length of string AC = 40√3 The inclination of string with the ground is 60∘ So, the required triangle is as follows sin60∘ = h40√3 √32 = h40√3 h = 60 m

A kite is flying on a string of length 40 $\sqrt{3}$ m. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^{\circ}$ . Find the height at which the kite is flying.

The correct option is B
60 m

Length of string AC = 40 $\sqrt{3}$

The inclination of string with the ground is $60^{\circ}$

So, the required triangle is as follows

$s i n 60^{\circ}$ = $\frac{h}{40 \sqrt{3}}$

$\frac{\sqrt{3}}{2}$ = $\frac{h}{40 \sqrt{3}}$

h = 60 m