Question

An aeroplane flying horizontal $1$ 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

Let $AC$ be the height of the aeroplane above the ground

In $△ABC,$

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

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

$\therefore$ $x=\frac{1}{\sqrt{3}}$ $...\left(1\right)$

Now, In $△BED,$

$\tan {30}^o=\frac{ED}{BD}$

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

$\Rightarrow$ $1=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}+y}$ [From $\left(1\right)$]

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

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

$\Rightarrow$ $Speed=\frac{Distance}{Time}$

$=\frac{\frac{2}{\sqrt{3}}km}{10seconds}$

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

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

$=\frac{720}{\sqrt{3}}$ $(Take\sqrt{3}=1.73)$

$=415.69\frac{km}{hr}$