Question

The angle of elevation of a jet plane from a point A on the ground is $60^{\circ }$. After a flight of 20 seconds at the speed of $432km/$ hour, the angle of elevation changes to $30^{\circ }.$ If the jet plane is flying at a constant height, then its height is :

• A. $1800\sqrt{3}m$
• B. $3600\sqrt{3}m$
• C. $2400\sqrt{3}m$
• D. $1200\sqrt{3}m$

Answer 1

Answer: (D)

$\tan 60^{\circ }=\frac{h}{y}$
$\sqrt{3}=\frac{h}{y}\Rightarrow h=\sqrt{3}y\dots \dots (1)$
$\tan 30^{\circ }=\frac{h}{x+y}$
$\frac{1}{\sqrt{3}}=\frac{h}{x+y}$
$\Rightarrow \sqrt{3}h=x+y\dots \dots (2)$
Speed $432km/h\Rightarrow \frac{432\times 20}{60\times 60}$
$\Rightarrow \frac{12}{5}km$
$\sqrt{3}h=\frac{12}{5}+y$
$\sqrt{3}h-\frac{12}{5}=y$
from (1)
$h=\sqrt{3}\left[\sqrt{3}h-\frac{12}{5}\right]$
$h=3h-\frac{12\sqrt{3}}{5}$
$h=\frac{6\sqrt{3}}{5}km$
$h=1200\sqrt{3}m$