Question

An aeroplane flying at a height of $9000\text{m}$ vertically above from the ground, passes another aeroplane when the angles of elevation of the two aeroplanes from a point on the ground are ${60}^{\circ }$ and ${30}^{\circ }$ respectively. Then the vertical distance between the aeroplanes at that instant is

• A. $1000\text{m}$
• B. $3000\text{m}$
• C. $6000\text{m}$
• D. $8000\text{m}$

Answer 1

The correct option is C $6000\text{m}$

Let the upper and lower aeroplanes be at point $B$ and $D$ respectively.


From $△OAB,$

$\tan {60}^{\circ }=\frac{9000}{x}$

$\Rightarrow \sqrt{3}=\frac{9000}{x}$

$\Rightarrow x=\frac{9000}{\sqrt{3}}\cdots \left(1\right)$

Now, from $△OAD,$

$\tan {30}^{\circ }=\frac{h}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}$

$\Rightarrow h=\frac{x}{\sqrt{3}}$

$\Rightarrow h=\frac{\frac{9000}{\sqrt{3}}}{\sqrt{3}}\left[\text{From}\right(1\left)\right]$

$\Rightarrow h=\frac{9000}{\sqrt{3}\times \sqrt{3}}$

$\Rightarrow h=\frac{9000}{3}$

$\Rightarrow h=3000\text{m}$

Hence, the vertical distance between both the planes is, $9000-3000=6000\text{m}$.