**Given: **Angle of depression of two consecutive stones when observed from an aeroplane are 30 degree and 60 degree.

We have to find the height at which the aeroplane is flying.

Consider,

Let, AB = h km

BC = x km and CD = 1km

Consider, right angled triangle ABC,

tan60 = AB/BC

=> \(\sqrt{3} = \frac{h}{x}\)

=> \(x = \frac{h}{\sqrt{3}}\) .....(i)

In right angled triangle ABD, we have

tan30 = AB/BD

=> \(\frac{1}{\sqrt3} = \frac{h}{x+1}\)

=> \(x+1 = \sqrt{3}h\) ....(ii)

From(i) and (ii), we get

\(\frac{h}{\sqrt{3}} + 1 = \sqrt{3}h\)

=> \(h+\sqrt{3} = 3h\)

=> \(2h = \sqrt{3}\)

=> So, the height at which the aeroplane is flying is \(h = \frac{\sqrt{3}}{2} km\)