An aeroplane flying at a constant speed, parallel to the horizontal ground, √3 km above it, is observed at an elevation of 60° from a point on the ground. If, after five seconds, its elevation from the same point is 30°, then the speed (in km/hr) of the aeroplane is?

Let the observing point be O.
The height at which the plane was flying above ground is constant at AP = QB = √3 km.
In the first case, distance of projection of plane from point of observation is tan60° = AP/OP => OP = 1 km.
In the second case, distance of projection of plane from point of observation is tan30° = BP/OQ => OQ = 3 km.
So, a distance of 3 - 1 = 2 km is covered in 5 seconds.
So the speed of the plane is (2 × 3600)/5 = 1440 km/hr.
So option D is the correct answer.