Question

The angle of elevation of an aeroplane from a point $A$ on the ground is ${60}^{\circ }$. After a flight of $15$ seconds horizontally, the angle of elevation changes to ${30}^{\circ }$. If the aeroplane is flying at a speed of $200$ m/s, then find the constant height at which the aeroplane is flying.

Answer 1

Let $A$ be the point of observation.
Let $E$ and $D$ be positions of the aeroplane initially and after $15$ seconds respectively.
Let $BE$ and $CD$ denote the constant height at which the aeroplane is flying.
Given that $\angle DAC={30}^{\circ },\angle EAB={60}^{\circ }$.
Let $BE=CD=h$ metres.
The distance covered in $15$ seconds,
$ED=200\times 15=3000$m ( distance travelled = speed $\times$ time )
Thus, $BC=3000$m.
In the right angled $△DAC$,
$\tan {30}^{\circ }=\frac{CD}{AC}$
$\Rightarrow CD=AC\tan {30}^{\circ }$
Thus, $h=(x+3000)\frac{1}{\sqrt{3}}$. (1)
In the right angled $△EAB$,
$\tan {60}^{\circ }=\frac{BE}{AB}$
$\Rightarrow BE=AB\tan {60}^{\circ }\Rightarrow h=\sqrt{3}x$ (2)
From (1) and (2), we have $\Rightarrow \sqrt{3}x=\frac{1}{\sqrt{3}}(x+3000)$
$\Rightarrow 3x=x+3000\Rightarrow x=1500$.
Thus, from (2) it follows that $h=1500\sqrt{3}$m.
The constant height at which the aeroplane is flying, is $1500\sqrt{3}m$.