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 km/hr.

Answer 1

An aero plane is flying 1 km above the ground making an angle of elevation of aero plane 60°. After 10 seconds angle of elevation is changed to 30°. Let CE = x, BC = y, $\angle AEB=30,\angle DEC=60,AB=1kmandCD=1km$ Here we have to find speed of aero plane.

The corresponding figure is as follows

So we use trigonometric ratios.

$In△DCE$,

$tan60=\frac{1}{x}$

$\sqrt{3}=\frac{1}{x}$

$x=\frac{1}{\sqrt{3}}$

$Againin△ABE$,

$tan30=\frac{1}{x+y}$

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

$x+y=\sqrt{3}$

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

$y=\sqrt{3}-\frac{1}{\sqrt{3}}$

$y=\frac{2}{\sqrt{3}}$

$speed=\frac{distance}{time}$

$=\frac{y}{10sec}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{10}{60\times 60}}$

$=425.68$

Hence the speed of aero plane is 415.68km/h.