Question

The angle of elevation of an aeroplane from a point on the ground is ${60}^o$. After flying for $15$ seconds, the elevation changes to ${30}^o$. If the aeroplane is flying at a speed of $720kmph$, then find the constant height at which aeroplane is flying?

Answer 1

REF.Image

distance covered in 15 seconds (y) =

$speed\times time$

$=720\times \frac{1000}{3600}\times 15=3000$ mtrs.

y = 3 km.

$tan{60}^{\circ }=\sqrt{3}=\frac{x}{z}\Rightarrow x=\sqrt{3}z\Rightarrow z=\frac{x}{\sqrt{3}}$

$tan{30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{x}{z+3}\Rightarrow z+3=\sqrt{3}x$

$z=\sqrt{3}x-3$

$\therefore$ Equating value -

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

$\Rightarrow 2x=3\sqrt{3}\Rightarrow x=\frac{3\sqrt{3}}{2}km$

$x=2.598km$