An aeroplane is flying horizontally at a height of 1.8 km above the ground. The angle of elevation of plane from point X is 60° and after 20 seconds, its angle of elevation from X is become 30°. If point X is on ground, then what is the speed (in m/s) of aeroplane?

In ∆ABC,
cot 60° = AC/BC
=> 1/√3 = AC/1800
=> AC = 1800/√3 ....(i)

In ∆AED,
cot 30° = AE/ED
=> √3 = AE/1800
=> AE = 1800√3 ....(ii)

From figure,
BD = CE
and, CE = AE - AC
so, BD = AE - AC

Now, substituting value of AE and AC from equations (i) and (ii),
BD = 1800√3 - 1800/√3
= 1800 (√3 - 1/√3)
= 1800 × 2/√3
BD = 3600/√3

BD is covered by the aeroplane in 20 seconds. So,
Speed of Aeroplane = Distance (BD) ÷ Time (t)
= (3600/√3 ÷ 20) m/sec
= 180√3 m/sec

A is correct choice.