Question

An airplane when flying at a height of $2000m$ from the ground passes vertically above the helicopter at an instant when the angles of the elevation of the two planes from the same point on the ground are ${60}^{\circ }$ and ${30}^{\circ }$ respectively. Find the vertical distance between the helicopter from the ground.

Answer 1

Let the distance between the two planes $A$ and $B$ be $x.$

In $△ADC,$

$\begin{array}{cc}\tan {60}^{\circ }& =\frac{2000}{y}\\ \sqrt{3}& =\frac{2000}{y}\dots \left(i\right)\end{array}$

Now, in $△BCD,$

$\begin{array}{cc}\tan {30}^{\circ }& =\frac{2000-x}{y}\\ \frac{1}{\sqrt{3}}& =\frac{2000-x}{y}\dots \left(ii\right)\end{array}$

Divide equation $\left(i\right)$ from $\left(ii\right),$

$\begin{array}{cc}3& =\frac{2000}{2000-x}\\ 6000-3x& =2000\\ 3x& =4000\\ x& =\frac{4000}{3}m\end{array}$

$\begin{array}{cc}2000-x& =2000-\frac{4000}{3}\\ & =\frac{6000-4000}{3}\\ & =\frac{2000}{3}m\end{array}$

Hence, the vertical distance of helicopter from the ground is equal to $\frac{2000}{3}m.$