A kite is flying at a height of 60 m above the ground. 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 length of the string, assuming that there is no slack in the string.

Let A be the point where kite is present, AC is the string and C is the point where string is tied on the ground.

In △ABC,

sin 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{60}{AC} \\[1em] \Rightarrow AC = \dfrac{60 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{120}{\sqrt{3}}$

Multiplying numerator and denominator by $\sqrt{3}$,

$\Rightarrow AC = \dfrac{120}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{120\sqrt{3}}{3} \\[1em] \Rightarrow AC = 40\sqrt{3} \text{ m}.$

Hence, length of string = $40\sqrt{3}$ m.