Question

An aeroplane flying with uniform speed horizontally $1$ km above the ground is observed at an elevation of ${60}^{\circ }$ from a point on the ground. After $10$ seconds, if the elevation is observed to be ${30}^{\circ },$ then the speed of the plane $($in km/hr$)$ is

• A. $\frac{240}{\sqrt{3}}$
• B. $200\sqrt{3}$
• C. $240\sqrt{3}$
• D. $\frac{120}{\sqrt{3}}$

Answer 1

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


Let $O$ be the point of observation and $A$ be the position of the aeroplane such that $\angle AOC={60}^{\circ }$ and $AC=1$ km.
After $10$ seconds, let $B$ be the position of the aeroplane such that $\angle BOD={30}^{\circ }$ and $BD=1$ km

In right angled triangle $AOC,$
$\tan {60}^{\circ }=\frac{AC}{OC}$
$\Rightarrow \sqrt{3}=\frac{1}{OC}$
$\Rightarrow OC=\frac{1}{\sqrt{3}}$

In right angled triangle $BOD,$
$\tan {30}^{\circ }=\frac{BD}{OD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{OD}$
$\Rightarrow OD=\sqrt{3}$

Now, $CD=OD-OC$
$=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$
Distance covered by the aeroplane in $10$ seconds $=\frac{2}{\sqrt{3}}$ km
Time taken $=10$ sec $=\frac{10}{3600}=\frac{1}{360}$ hr
Speed of the aeroplane
$=\frac{\text{distance}}{\text{time}}=\frac{2/\sqrt{3}}{1/360}=\frac{720\sqrt{3}}{3}$
$=240\sqrt{3}$ km/hr