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# 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 $60^o$. Find the length of the string, assuming that there is no slack in the string.

Given:

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 $60^o$.

To do:

We have to find the length of the string, assuming that there is no slack in the string.

Solution:

Let $AB$ be the height above ground and $AC$ be the length of the string of the kite.

From the figure,

$\mathrm{AB}=60 \mathrm{~m}, \angle \mathrm{ACB}=60^{\circ}$

Let the length of the string be $\mathrm{AC}=h \mathrm{~m}$

We know that,

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{\text { AB }}{AC}$

$\Rightarrow \sin 60^{\circ}=\frac{60}{h}$

$\Rightarrow \frac{\sqrt3}{2}=\frac{60}{h}$

$\Rightarrow h=60 \times \frac{2}{\sqrt3} \mathrm{~m}$

$\Rightarrow h=\frac{120}{\sqrt3} \mathrm{~m}$

$\Rightarrow h=\frac{120\sqrt3}{(\sqrt3)^2} \mathrm{~m}$

$\Rightarrow h=\frac{120\times\sqrt3}{3} \mathrm{~m}$

$\Rightarrow h=40\sqrt3 \mathrm{~m}$

Therefore, the length of the string is $40\sqrt3 \mathrm{~m}$.

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