# The angle of elevation of a jet plane from a point A

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 $432 \mathrm{~km} /$ hour, the angle of elevation changes to $30^{\circ}$. If the jet plane is flying at a constant height, then its height is :

1. $1800 \sqrt{3} \mathrm{~m}$

2. $3600 \sqrt{3} \mathrm{~m}$

3. $2400 \sqrt{3} \mathrm{~m}$

4. $1200 \sqrt{3} \mathrm{~m}$

Correct Option: , 4

Solution:

$\tan 60^{\circ}=\frac{h}{y}$

$\sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y$ .......................(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)

$\mathrm{h}=\sqrt{3}\left[\sqrt{3} \mathrm{~h}-\frac{12}{5}\right]$

$\mathrm{h}=3 \mathrm{~h}-\frac{12 \sqrt{3}}{5}$

$\mathrm{~h}=\frac{6 \sqrt{3}}{5} \mathrm{~km}$

$\mathrm{h}=1200 \sqrt{3} \mathrm{~m}$