An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} \mathrm{~km}$ above it…
An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} \mathrm{~km}$ above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$, then the speed (in $\mathrm{km} / \mathrm{hr}$ ) of the aeroplane is
1500
750
720
1440
Solution
For $\Delta \mathrm{OA}, \mathrm{A}, \mathrm{OA}_1=\frac{\sqrt{3}}{\tan 60^{\circ}}=1 \mathrm{~km}$
For $\Delta \mathrm{OB}_1, \mathrm{~B}, \mathrm{OB}_1=\frac{\sqrt{3}}{\tan 30^{\circ}}=3 \mathrm{~km}$.
As, a distance of $3-1=2 \mathrm{~km}$ is covered in 5 seconds. Therefore the speed of the plane is
$
\frac{2 \times 3600}{5}=1440 \mathrm{~km} / \mathrm{hr}
$
