An aeroplane flying horizontally at a height of 3 Km. above the ground is observed at a certain point on earth to subtend an angle of 60°. After 15 sec flight, its angle of elevation is changed to 30°. The speed of the aeroplane (taking √3 = 1.732) is

AB = CD = 3000 m

A and C are positions of aeroplane.

$$\angle AOB$$

$$\angle AOB = 60\degree      \angle COD = 30\degree$$

In $$\triangle OAB$$,

$$\tan 60\degree = \frac{AB}{OB}$$

=> $$\sqrt{3} = \frac{3000}{OB}$$

=>OB = $$\frac{3000}{\sqrt{3}} = 1000\sqrt{3}$$

In $$\triangle OCD$$,

$$\tan 30\degree = \frac{CD}{OD}$$

=> $$\frac{1}{\sqrt{3}} = \frac{3000}{OD}$$

=>CD = $$3000\sqrt{3} = 3000\sqrt{3}$$

BD = $$3000\sqrt{3} - 1000\sqrt{3} = 2000\sqrt{3}$$

Speed of aeroplane 

=$$\frac{2000\sqrt{3}}{15}$$

= 230.93

So, the answer would be option b)230.93 m/sec.