Question

An aeroplane (A) when flying at a certain height from the ground passes vertically above another aeroplane (B) at an instant when the angles of the elevation of the two aeroplanes from the same point on the ground are 30º and 45º respectively and the distance between two aeroplanes is 1000 m. Then, the approximate vertical height of the aeroplane A is $(\text{U}se\sqrt{3}=1.732)$

• A. 2500 m
• B. 2366 m
• C. 1183 m
• D. 1300 m

Answer 1

The correct option is B 2366 m

Let the point on the ground from where angles of elevation of aeroplanes A and B is taken, be C.
Let the perpendicular dropped on the ground from A and B be D.
Let BD = x.

$In\Delta CBD,$
$\tan {30}^{\circ }=\frac{BD}{CD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{CD}$
$\Rightarrow CD=\sqrt{3}x$
$In\Delta CAD,$
$\tan {45}^{\circ }=\frac{AD}{CD}$
$\Rightarrow 1=\frac{x+1000}{\sqrt{3}x}$ $(\because AD=AB+BD)$
$\Rightarrow \sqrt{3}x=x+1000$
$\Rightarrow 1.732x-x=1000$
$\Rightarrow 0.732x=1000$
$\Rightarrow x=\frac{1000}{0.732}=1366m$
∴ AD = 1000 + x = 1000 + 1366 = 2366 m
Thus, the approximate vertical height of the aeroplane A is 2366 m.
Hence, the correct answer is option (b).