Question

The angle of elevation of an aeroplane from point A on the ground is ${60}^o$. After flight of $15$ seconds, the angle of elevation changes to ${30}^o$. 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

Height of aeroplane $=BD=CE=1500\sqrt{3}m$ and

$\angle BAE={60}^0$ and $CAE={30}^0$

In triangle $ADB$

$\tan {60}^0=\frac{1500\sqrt{3}}{AD}$

$\Rightarrow \sqrt{3}=\frac{1500\sqrt{3}}{AD}$

$\Rightarrow AD=1500m$

In triangle $CAE$

$\tan {30}^0=\frac{1500\sqrt{3}}{AE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AD}$

$\Rightarrow AE=1500\times 3=4500m$

Distance covered by plane in $15$ seconds: $BC=DE=AE-AD=4500-1500=3000m$

Speed of aeroplane $=\frac{3000}{15}=200m/s=720km/hr$