A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 30° and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60°. If the bird flies all along in a straight line at a height of 50 m, then its speed in km/h is :

From the diagram,
Height = AD = 50√3 m
∠BAN = 30°
∠CAM = 60°
∴∠BAD = 90° - 30° = 60°
∴∠CAD = 90° - 60° = 30°
From ΔABD,
tan∠BAD = Perpendicular/ Base
tan60° = BD/AD
√3 = BD/(50√3)
BD = 50 × 3 = 150 m
From ΔACD,
tan∠CAD = Perpendicular/ Base
tan30° = CD/AD
1/√3 = CD/(50√3)
CD = 50 m
∴ Distance travelled by the bird
= BC = BD + CD = 150 m + 50 m = 200 m = 0.200 km
Time taken to cover this distance = 2 minutes = 2/60 hr = 1/30 hr
∴ Speed
= Distance travelled/ Time required
=0.200 $$\times$$30km/hr
= 0.200 × 30 km/hr
= 6 km/hr
Option D is the correct answer