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 $432 \mathrm{~km} /$ hour, the angle of elevation changes to $30^{\circ}$. If the jet plane is flying at a constant height, then its height is :
Correct Option: , 4
$\tan 60^{\circ}=\frac{h}{y}$
$\sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y$ .......................(1)
$\tan 30^{\circ}=\frac{h}{x+y}$
$\frac{1}{\sqrt{3}}=\frac{h}{x+y} \Rightarrow \sqrt{3} h=x+y$ ......................(2)
Speed $432 \mathrm{~km} / \mathrm{h} \Rightarrow \frac{432 \times 20}{60 \times 60} \Rightarrow \frac{12}{5} \mathrm{~km}$
$\sqrt{3 h}=\frac{12}{5}+y$
$\sqrt{3} h-\frac{12}{5}=y$
from (1)
$\mathrm{h}=\sqrt{3}\left[\sqrt{3} \mathrm{~h}-\frac{12}{5}\right]$
$\mathrm{h}=3 \mathrm{~h}-\frac{12 \sqrt{3}}{5}$
$\mathrm{~h}=\frac{6 \sqrt{3}}{5} \mathrm{~km}$
$\mathrm{h}=1200 \sqrt{3} \mathrm{~m}$