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 a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

Solution:

Let angle of elevation of an aero plane is 45°. After 15 second angle of elevation is change to 30°. Let DE be the height of aero plane which is 3000 meter above the ground. Let  and. Here we have to find speed of aero plane.

We have the corresponding figure as follows

So we use trigonometric ratios.

In $\triangle A B C$

$\Rightarrow \quad \tan A=\frac{B C}{A B}$

$\Rightarrow \quad \tan 45^{\circ}=\frac{3000}{x}$

$\Rightarrow \quad 1=\frac{3000}{x}$

$\Rightarrow \quad x=3000$

Again in $\triangle A D E$

$\Rightarrow \quad \tan A=\frac{D E}{A B+B D}$

$\Rightarrow \quad \tan 30^{\circ}=\frac{3000}{x+y}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{3000}{3000+y}$

 

$\Rightarrow \quad 3000+y=3000 \sqrt{3}$

$\Rightarrow \quad y=3000 \sqrt{3}-3000$

$\Rightarrow \quad y=3000(\sqrt{3}-1)$

$\Rightarrow \quad y=2196$

Since $15 \mathrm{sec}=2196$

$\Rightarrow \quad \mathrm{sec}=\frac{2196}{15}=146.4$

$\Rightarrow \quad \sec =\frac{2196}{15}=146.4$

$=\frac{146.4 \times 3600}{1000}$

$=527.04$

Hence the speed of aero plane is $527.04 \mathrm{~km} / \mathrm{h}$.

 

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