Question

An aeroplane when flying at a height of $5000m$ ffrom the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are ${60}^0$ and ${45}^0$ respectively. Find the vertical distance between the aero planes at the instant.

Answer 1

Let $AB$ be th horizontal distance, the first aeroplane is flying at point $C$ and second aeroplane is flying at point $D$

In $△ACB$

$\tan {60}^0=\frac{5000}{x}$

$x=\frac{5000}{\sqrt{3}}$

In $△ADB$

$\tan {45}^0=\frac{y}{x}=1$

$\Rightarrow y=x=\frac{5000}{\sqrt{3}}$

$\therefore$Vertical distance between planes

$CD=CB-DB=5000-\frac{5000}{\sqrt{3}}$

$=5000\left[\frac{\sqrt{3}-1}{\sqrt{3}}\right]$

Taking $\sqrt{3}=1.732$, we get

$CD=2113.164$