Question

The angle of elevation of an aeroplane from a point on the ground is ${45}^{\circ }$. After a flight of $15$ sec the elevation changes to ${30}^{\circ }$. If the aeroplane is flying at a height of $3000$ metres, find the speed of the aeroplane.

• A. $304$ km/hr
• B. $457$ km/hr
• C. $321.37$ km/hr
• D. $527.04$ km/hr

Answer 1

Answer: (D)

$527.04$ km/hr

Let the point in the ground is $E$ which is y metres from point $B$ and let after $15$ sec flight, it covers $x$ metres distance
In $\Delta AEB$
$\tan {45}^{\circ }=\frac{AB}{EB}$

$\therefore 1=\frac{3000}{y}$

$\therefore y=3000m$ .......(i)

In $\Delta CED$
$\tan {30}^{\circ }=\frac{CD}{ED}$

$\therefore \frac{1}{\sqrt{3}}=\frac{3000}{x+y}$ ($\because$ $AB=CD$)

$\therefore x+y=3000\sqrt{3}$ .......(ii)

From equation (i) and (ii)
$x+3000=3000\sqrt{3}$
$\therefore x=3000\sqrt{3}-3000$
$\therefore x=3000(\sqrt{3}-1)$
$\therefore$ $x=3000(1.732-1)$
$\therefore$ $x=3000\left(0.732\right)$
$\therefore$ $x=2196m$

Speed of Aeroplane = $\frac{\text{Distance covered}}{\text{Time taken}}$
$=\frac{2196}{15}$ m/sec = 146.4 m/sec

$=\frac{2196}{15}\times \frac{18}{5}$ km/hr

$=527.04$km/hr
Hence the speed of aeroplane is $527.04$ km/hr