Question

A balloon is connected to a meteorological ground station by a cable of 215 m inclined at $60{}^\circ$ to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.

Answer 1

Hint: For solving this question, we will first draw the figure for the given situation and then use one of the trigonometric ratios, that is the sine ratios. Sine ratio is defined as the ratio of perpendicular to hypotenuse, that is, $\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$.

Complete step-by-step answer:
In this question, we have been asked to find the height of a balloon when it is tied with a cable of 215 m at an angle of $60{}^\circ$. So, from the given conditions, we can draw the figure as given below.

Here, we have considered the position of the balloon as point B, which is tied with a cable whose other end is at point A. So, we can write the length of the cable as same as AB = 215 m. Also, we have been given that the cable is connected to the balloon at an inclination of $60{}^\circ$ to the horizontal surface. So, we get $\angle BAC=60{}^\circ$. Now we know that the sine ratio of trigonometry is defined as the ratio of perpendicular to hypotenuse, that is, $\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$. So, we can write,
For $\Delta ABC,\sin \angle BAC=\dfrac{BC}{AB}$
Now we will put the values of $\angle BAC$ and AB, that is, $\angle BAC=60{}^\circ$ and AB = 215 m. So, we get,
$\sin 60{}^\circ =\dfrac{BC}{215}$
Now, we know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$. So, we will substitute the same, so we get,
$\begin{aligned} & \dfrac{\sqrt{3}}{2}=\dfrac{BC}{215} \\\ & \Rightarrow \dfrac{\sqrt{3}}{2}\times 215=BC \\\ & \Rightarrow BC=\left( 107.5 \right)\sqrt{3} \\\ \end{aligned}$
Hence, we can say that the height of the balloon is $107.5\sqrt{3}$ metres.

Note: In this question, there are high possibilities that we may make silly mistakes by writing the wrong value of $\sin 60{}^\circ$ or we may also take the reciprocal of $\dfrac{BC}{AB}$ as the sine ratio, which would be wrong. So, we have to remember that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$.