Question

A kite is flying at a height of 60m 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 spring.\[\]
A. $l=60\sqrt{2}\text{m}$\[\]
B. $l=40\sqrt{3}\text{m}$\[\]
C. $l=20\sqrt{3}\text{m}$\[\]
D. None of these \[\]

Answer 1

Hint: We denote the position of the kite with the point A, the foot of perpendicular drawn from A to the ground as B, and the position of the point where the string is tied as C. We are given $AB=60\text{m},\angle ACB={{60}^{\circ }}$. We take the sine of $\angle ACB$ and find the length of the string $l=AC$.\[\]

Complete step-by-step solution:
We know from the trigonometry that in a right-angled triangle the sine of an acute angle is given by the ratio of the side opposite to the said angle to the hypotenuse
We are given the question that a kite is flying at a height of 60m above the ground. Let us denote the position of the kite as A and we drop perpendicularly from A which meets the ground on B. So the height of the kite is
\[AB=60\text{m}\]
We are further given in the question that the string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is ${{60}^{\circ }}$. Let that point be C. So the angle of inclination is
\[\angle ACB={{60}^{\circ }}\]
We are also given in the question that there is no slack in the spring. So the length of the spring is equal to the distance between A and C that is $AC$ which are asked to find. We draw the rough figure below.
 We take the sine of the angle of $\angle ACB$ (whose opposite side is $AB$) in right angled triangle ABC (whose hypotenuse is side $AC$) to have,
\[\begin{align}
  & \sin \left( \angle ACB \right)=\dfrac{AB}{AC} \\
 & \Rightarrow \sin \left( {{60}^{\circ }} \right)=\dfrac{60}{AC} \\
&\Rightarrow AC=\dfrac{60}{\sin {{60}^{\circ }}}=\dfrac{60}{\dfrac{\sqrt{3}}{2}}=\dfrac{120}{\sqrt{3}}=\dfrac{120\sqrt{3}}{3}=40\sqrt{3} \\
\end{align}\]
So the length of the string is $l=40\sqrt{3}$m and the correct option is B.\[\]


Note: We must be careful of the confusion between the values of $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ because if we put $\dfrac{1}{\sqrt{2}}$ instead of $\dfrac{\sqrt{3}}{2}$ we will get the incorrect option A. We can find the side BC by taking the tangent of $\angle ACB$. If we observe the kite from the point C then$\angle ACB$ becomes the angle of elevation.