An aeroplane 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° and 45° respectively. Find the vertical distance between the two planes.

Let the plane 3000 m high be at point B and plane below it be at point D.

From figure,

Considering right angled triangle △ABC,

$\Rightarrow \text{tan 60°} = \dfrac{BC}{AC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{3000}{AC} \\[1em] \Rightarrow AC = \dfrac{3000}{\sqrt{3}} \\[1em] \Rightarrow AC = 1732 \text{ m}.$

Considering right angled triangle △ADC,

$\Rightarrow \text{tan 45°} = \dfrac{DC}{AC} \\[1em] \Rightarrow 1 = \dfrac{DC}{1732} \\[1em] \Rightarrow DC = 1732 \text{ m}.$

Distance between two planes (BD) = BC - DC = 3000 - 1732 = 1268 m.

Hence, the distance between two planes is 1268 m.