An aeroplane flying horizontally 1 km above the ground is observed...

Let, $d$ is the distance covered by aeroplane in $10$ sec.

Let, the height at which the plane is flying be $H = 1km$

$⇒ H = AC = BD = 1km$

In $∆ OAC$,
$tan {60}^{o} = \dfrac{AC}{OC}$

$OC = 1cot {60}^{o}$...(i)

In $∆ OBD$,
$tan {30}^{o} = \dfrac{BD}{OD}$

$OD = 1cot {30}^{o}$...(ii)
Now,
$d = AB = CD = OD - OC$
$d = cot {30}^{o} - cot {60}^{o}$

Speed = \dfrac{d}{t} = \dfrac{cot {30}^{o} - cot {60}^{o}}{10}km / sec$<br /><br />Speed in$km / h$<br /><br />$= \dfrac{cot {30}^{o} - cot {60}^{o}}{10} \times 60 \times 60$<br /><br />$= \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{10} \times 3600$<br /><br />$= 240\sqrt{3}km / h