Question

An aeroplane when flying at a height of $3125m$ from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are ${30}^o$ and ${60}^o$ respectively. Find the distance between the two planes at that instant.

Answer 1

In $△BCD,$

$\Rightarrow \tan 30°=\frac{BC}{CD}=\frac{3125}{CD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3125}{CD}$

$\Rightarrow CD=3125\sqrt{3}m⟶\left(1\right)$

In $△ADC,$

$\Rightarrow \tan 60°=\frac{AC}{CD}=\frac{AB+BC}{CD}$

$\Rightarrow \sqrt{3}=\frac{AB+3125}{CD}$

$\Rightarrow CD=\frac{AB+3125}{\sqrt{3}}⟶\left(2\right)$

From $\left(1\right)\&\left(2\right)$ we get

$\Rightarrow 3125\sqrt{3}=\frac{(AB+3125)}{\sqrt{3}}$

$\Rightarrow AB=(3125\times 3)-3125$

$\Rightarrow AB=3125\times 2$

$\Rightarrow AB=6250m$

$\therefore$ Distance between the two planes is $6250m$

Hence, the answer is $6250.$