Question

An aeroplane flying horizontally $1$km above the ground is observed at an elevation of ${60}^{\circ }$.After $10$ seconds, its elevation is observed to be ${30}^{\circ }$.Find the speed of the aeroplane in kmph.

Answer 1

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,$△ODB,$

$\tan {30}^{\circ }=\frac{BD}{OD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{OD}$

$\Rightarrow OD=\sqrt{3}$

Now, $CD=OD-OC=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{3-1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

Distance covered by the aeroplane in $10$ seconds$=\frac{2}{\sqrt{3}}$km

Time taken$=10$secs$=\frac{10}{3600}=\frac{1}{360}$hrs

Speed of the aeroplane$=\frac{Distance}{Time}=\frac{\frac{2}{\sqrt{3}}}{\frac{1}{360}}=\frac{720\sqrt{3}}{3}=240\times 1.732=415.68$km/hr