An aeroplane when flying at a height of 3125 m 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

Question

An aeroplane when flying at a height of 3125 m 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 $30°$ and $60°$ respectively. Find the distance between the two planes at that instant.

Accepted Answer

Step 1: Let the aeroplanes be at positions $C$ and $D$.
Let $CD = x$, $AB = y$
$\angle BAC = 30°$, $\angle BAD = 60°$
Step 2: In right angled $	riangle ABC$, $	an 30° = \frac{3125}{y}$
$\frac{1}{\sqrt{3}} = \frac{3125}{y} \Rightarrow y = 3125\sqrt{3}$ m
Step 3: In right angled $	riangle ABD$, $	an 60° = \frac{x + 3125}{y}$
$\sqrt{3}y = x + 3125$
$\sqrt{3}(3125\sqrt{3}) = (x + 3125) \Rightarrow x = 2(3125)$
$x = 6250$ m
$	herefore$ Distance between two planes $= 6250$ m

Question

Solution