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 the instant.

Let the aeroplane at a height of 3125 m be at a point R and the plane above be at point S.

Considering right angled △PQR, we get

$\Rightarrow \text{tan 30°} = \dfrac{QR}{PQ} \\[1em] \Rightarrow \text{tan 30°} = \dfrac{3125}{PQ} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{3125}{PQ} \\[1em] \Rightarrow PQ = 3125\sqrt{3}$

Now considering right angled △PQS, we get

$\Rightarrow \text{tan 60°} = \dfrac{QS}{PQ} \\[1em] \Rightarrow \sqrt{3} = \dfrac{QS}{3125\sqrt{3}} \\[1em] \Rightarrow QS = 3125\sqrt{3} \times \sqrt{3} \\[1em] \Rightarrow QS = 9375$

Distance between two aeroplanes = QS - QR = 9375 - 3125 = 6250 m.

Hence, the distance between two planes = 6250 m.