Question

An aeroplane flying horizontally 1 km. above the ground is observed at an elevation of ${60}^o$ and after $10$ seconds the elevation is observed to be ${30}^o$. The uniform speed of the aeroplane in $km/h$ is

• A. $240$
• B. $240\sqrt{3}$
• C. $60\sqrt{3}$
• D. None of these

Answer 1

The correct option is B $240\sqrt{3}$

In triangle with angle of elevation ${30}^{\circ }$
$\tan 30=\frac{1}{d_a}$
$d_a=\cot 30=\sqrt{3}$
In triangle with angle of elevation ${60}^{\circ }$
$\tan 60=\frac{1}{d_b}$
$d_b=\cot 60=\frac{1}{\sqrt{3}}$
Thus, $d=d_a-d_b=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$ km
Now $speed=\frac{distance}{time}$
$speed=\frac{\frac{2}{\sqrt{3}}}{10}\times 60\times 60$
$speed=240\sqrt{3}$ km/hr