An aeroplane when 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are \(60\degree\) and \(45\degree\) respectively. The vertical distance between the two aeroplanes is
In \(\triangle\)ABC tan 60°=\(\frac{3000}{BC}\) BC = \(\frac{3000}{\sqrt{3}}\)m ………..(i) In \(\triangle\)BCD tan 45°=\(\frac{3000 - h}{BC}\) BC = 3000 − h ……. (ii) Equating and solving equations (i) and (ii), we get h = 1268m So the correct option is (A).