Question

2. 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 degree and 45 degree respectively. find the vertical distance between the two planes.​

Answer 1

The vertical distance between two planes is 1268 m

Step-by-step explanation:

As shown in figure A and B are the positions of the aeroplanes. the angle of elevation of A is 60° and the angle of elevation of B is 45°

Let CD = x and BC = h

In ΔACD

$\tan 60^\circ=\frac{AC}{CD}$

$\implies \sqrt{3}=\frac{3000}{x}$

In ΔBCD

$\tan 45^\circ=\frac{BC}{CD}$

$\implies 1=\frac{h}{x}$

$\implies h=x$

Therefore,

$\sqrt{3}=\frac{3000}{h}$

$\implies h=\frac{3000}{\sqrt{3}}$

$\implies h=\frac{3000\sqrt{3}}{3}$

$\implies h=1000\times 1.732$

$\implies h=1732$ m

Therefore, the vertical distance between the planes

$=3000-1732$

$=1268$ m

Hope this answer is helpful.

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