# The angle of elevation of an aeroplane from a point on the ground is 45°.

Question:

The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane.

Solution:

Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

$\mathrm{PQ}=\mathrm{BC}=2500 \mathrm{~m}, \angle \mathrm{PAQ}=45^{\circ}$ and $\angle \mathrm{BAC}=30^{\circ}$

In $\Delta \mathrm{PAQ}$,

$\tan 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}$

$\Rightarrow 1=\frac{2500}{\mathrm{AQ}}$

$\Rightarrow \mathrm{AQ}=2500 \mathrm{~m}$

Also, in $\triangle \mathrm{ABC}$,

$\tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{2500}{\mathrm{AC}}$

$\Rightarrow \mathrm{AC}=2500 \sqrt{3} \mathrm{~m}$

Now, QC = AC - AQ

$=2500 \sqrt{3}-2500$

$=2500(\sqrt{3}-1) \mathrm{m}$

$=2500(1.732-1)$

$=2500(0.732)$

$=1830 \mathrm{~m}$

$\Rightarrow \mathrm{PB}=\mathrm{QC}=1830 \mathrm{~m}$

So, the speed of the aeroplane $=\frac{\mathrm{PB}}{15}$

$=\frac{1830}{15}$

$=122 \mathrm{~m} / \mathrm{s}$

$=122 \times \frac{3600}{1000} \mathrm{~km} / \mathrm{h}$

$=439.2 \mathrm{~km} / \mathrm{h}$

So, the speed of the aeroplane is 122 m/s or 439.2 km/h.