Question

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

• A. $240$
• B. $240\sqrt{3}$
• C. $60\sqrt{3}$
• D. $120$

Answer 1

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

In the above figure,
$AM=H=1\text{km},BC=d,\angle BAN={60}^{\circ },\angle CAN={30}^{\circ }\Rightarrow d=H\cot {30}^{\circ }-H\cot {60}^{\circ }$
Time taken $=10s$
So, speed $=\frac{d}{10}=\frac{H\cot {30}^{\circ }-H\cot {60}^{\circ }}{10}\times 60\times 60=240\sqrt{3}\text{km/h}$