$A$ 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 $60^{\circ}$. Find the length of the string,assuming that there is no slack in the string.

$(40\sqrt{3}\, m) $ Let $K$ be the position of the kite and $P$ be the point on the ground where the string is tied. Let $L$ be the point on the ground directly below the kite.
In the right-angled triangle $\triangle KLP$:
$KL = 60\, m$ (height of the kite)
$\angle KPL = 60^{\circ}$ (angle of inclination)
We need to find the length of the string,which is the hypotenuse $KP$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$:
$\sin 60^{\circ} = \frac{KL}{KP}$
$\frac{\sqrt{3}}{2} = \frac{60}{KP}$
$KP = \frac{60 \times 2}{\sqrt{3}}$
$KP = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$KP = \frac{120\sqrt{3}}{3} = 40\sqrt{3}\, m$
Thus,the length of the string is $40\sqrt{3}\, m$ (approximately $69.28\, m$).