Answer: Option D.
Let the plane at the start is at point C and after 15 seconds it reaches point D after covering x km as shown in diagram.
In Δ ABC, tan 45° = BC/AB => 1 = BC/AB => AB = BC = 3,000.
In %Delta; ADE, tan 30°= ED/(AB+BE) => 3000/(3000+BE) => BE = 2196 m .
Therefore x = 2.196 km.
So aeroplane covers 2.196 km in 15 seconds, so speed of aeroplane = (2.196 x 60 x 60)/15 = 527kmph.