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

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

Answer 1

The correct option is A $1200\sqrt{3}$ m


Velocity, $v=432\times \frac{1000}{60\times 60}m/sec=120m/sec$
Distance $PQ=v\times 20=2400m$
In $\Delta PAC$
$\tan {60}^{\circ }=\frac{h}{AC}\Rightarrow AC=\frac{h}{\sqrt{3}}$
In $\Delta AQD$
$\tan {30}^{\circ }=\frac{h}{AD}\Rightarrow AD=\sqrt{3}h$

Now, $AD=AC+CD$
$\Rightarrow \sqrt{3}h=\frac{h}{\sqrt{3}}+2400$
$\Rightarrow \frac{2h}{\sqrt{3}}=2400$
$\Rightarrow h=1200\sqrt{3}m$