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 $432 km /$ 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)

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