Question

An aeroplen when 3000 m high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation point are ${60}^{\circ }$ and ${45}^{\circ }$. How many meters higher is this one than the other.

• A. 1350 m
• B. 1268 m
• C. 1000 m
• D. 1160 m

Answer 1

The correct option is B 1268 m

Let the aeroplane at a height of $3000$ m at point A passes over another aeroplane at B at a height $x$ metres.
$\angle ADC={60}^{\circ }$
(Angle of elevation of higher aeroplane)
$\angle BDC={45}^{\circ }$
(Angle of elevation of lower aeroplane)
$\therefore In\Delta ADC,\frac{AC}{DC}=\tan {60}^{\circ }\Rightarrow \frac{3000}{DC}=\sqrt{3}\Rightarrow DC=\frac{3000}{\sqrt{3}}$ metres = $1732$ metres (approx)
$\therefore In\Delta BDC,\frac{BC}{DC}=\tan {45}^{\circ }\Rightarrow \frac{x}{1732}=1\Rightarrow x=1732m\therefore$ Difference in heights = $(3000-1732)$ metres
= $1268$ metres.