Question

An aeroplane when flying at a height 2500 m from the ground 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 ${45}^{\circ }$ and ${30}^{\circ }$ respectively then the vertical distance between the two aeroplanes at that instant is

• A. 1158.5 m
• B. 1056.5 m
• C. 1008.5 m
• D. None

Answer 1

The correct option is B 1056.5 m

Given
$\angle ACB={45}^{\circ }$
$\angle DCB={30}^{\circ }$
$AB=2500m$
$AD$ we have to find out.
From $△ABC$,
$\tan 45=\frac{AB}{BC}$
$BC=2500m$
From $△DBC$,
$\tan 30=\frac{DB}{BC}$
$DB=\frac{2500}{\sqrt{3}}$
$AD=AB-BD$
$=2500-\left(\frac{2500}{\sqrt{3}}\right)$
$=1056.5m$
So the vertical distance between the two aeroplanes at that instant is $=1056.5m$