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

\[\text { Let } \quad \begin{aligned} & C D=x \\ & A B=y \end{aligned} \] \[\angle B A C=30^{\circ}, \angle B A C=60^{\circ} \] In right angled \(\displaystyle \triangle A B C, \tan 30^{\circ}=\frac{3125}{y}\) \[\frac{1}{\sqrt{3}}=\frac{3125}{y} \Rightarrow y=3125 \sqrt{3} \mathrm{~m} \] In right angled \(\displaystyle \triangle A B D, \tan 60^{\circ}=\frac{x+3125}{y}\) \[\sqrt{3} y=x+3125 \] \[\sqrt{3}(3125 \sqrt{3})=(x+3125) \Rightarrow x=2(3125) \] \[x=6250 \mathrm{~m} \] ∴ Distance between two planes \(\displaystyle =6250 \mathrm{~m}\)