Question

The angle of elevation of an aeroplane from point A on the ground is ${60}^{\circ }$. After flight of 15 seconds, the angle of elevation change to ${30}^{\circ }$. If the aeroplane is flying at a constant height of $1500\sqrt{3}$ m, find the speed of the plane in km/hr.

Answer 1

Step1: Draw a diagram based on question.

Let BC be the height at which the aeroplane flying.

Then, BC = $1500\sqrt{3}m$

In 15 seconds,the aeroplane moves from C to E and makes angle of elevation

${30}^{\circ }$ at E.

Let AB = x m, BD = y m

So, AD = (x + y) m

Step2: Find the value of x

In $△ABC,$

$tan{60}^{\circ }=\frac{BC}{AB}$

$\sqrt{3}=\frac{1500\sqrt{3}}{x}[\because tan{60}^{\circ }=\sqrt{3}]$

$x=1500m$ ...(i)

Step3: Find the value of y

In $△EAD,$

$tan{30}^{\circ }=\frac{ED}{AD}$

$\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{x+y}[\because tan{30}^{\circ }=\frac{1}{\sqrt{3}}]$

$x+y=1500\times 3$

$y=4500-1500=3000m$ [$\because$ Using equation (i)]

Step4: Find the value of speed

Speed of aeroplane = $\frac{\text{Distance}}{\text{Time}}$

$=\frac{3000}{15}$

= 200 m/s
$=200\times \frac{18}{5}=720km/hr$.

Speed of aeroplane is 720 km/hr.