Question

An aeroplane flying at a constant speed, parallel to the horizonal ground, $\sqrt{3}$ km above it, is observed at an elevation of ${60}^o$ from a point on the ground. If, after five seconds, its elevation from the same point, is ${30}^o$, then the speed (in km/hr) of the aeroplane, is?

• A. $720$
• B. $1500$
• C. $750$
• D. $1440$

Answer 1

The correct option is D $1440$

An aeroplane is flying $\sqrt{3}km$ above the ground making an angle of elevation of aeroplane ${60}^{\circ }$.

After $5sec$ angle of elevation changed to ${30}^{\circ }$.

Let $CE=x,BC=y,\angle AEB={30}^{\circ },\angle DEC={60}^{\circ },AB=\sqrt{3}km$ and $CD=\sqrt{3}km$.

So, we now use trigonometric ratios,

In $△DCE$

$tan{60}^{\circ }=\frac{CD}{CE}$

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

$\Rightarrow x=1$

Again in $△ABE$, we have

$tan{30}^{\circ }=\frac{AB}{BE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{x+y}$

$\Rightarrow x+y=3$

$\Rightarrow 1+y=3[\because x=1]$

$\Rightarrow y=2$

Now, We know $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

$=\frac{y}{5sec}$

$=\frac{2}{\frac{5}{60\times 60hr}}$

$=1440$

Hence, the speed of aeroplane is $1440kmperhr$.

So, $\text{D}$ is the correct option.