The angle of elevation of an aeroplane from a point on the ground is $ 45^{\circ} $. After a flight of 15 seconds, the elevation changes to $ 30^{\circ} $. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.
Given:
The angle of elevation of an aeroplane from a point on the ground is \( 45^{\circ} \). After a flight of 15 seconds, the elevation changes to \( 30^{\circ} \).
The aeroplane is flying at a height of 3000 metres.
To do:
We have to find the speed of the aeroplane.
Solution:
Let $C$ be the plane flying in the sky at a height of $3000\ m$. After a flight of 15 seconds plane is at point $E$.
From the figure,
$BC=DE=3000\ m$
Let the distance between $A$ and $B$ be $y\ m$ and the distance between the points $C$ and $E$ be $x\ m$.
This implies,
$BD=CE=x\ m$
In right $\Delta \mathrm{CAB}$,
$\tan 45^{\circ}=\frac{CB}{AB}$
$1=\frac{3000}{y}$
$y=3000\ m$..........(i)
Similarly,
In right $\Delta \mathrm{EAD}$,
$\tan 30^{\circ}=\frac{ED}{AD}$
$\frac{1}{\sqrt3}=\frac{3000}{x+y}$
$x+y=3000\sqrt3\ m$
$\Rightarrow x=3000\sqrt3-3000 \mathrm{~m}$ [From(i)]
$\Rightarrow x=3000(\sqrt{3}-1)$
$=3000 \times(1.732-1)$
$=3000 \times 0.732$
$=2196 \mathrm{~m}$
Therefore,
The distance \( 2196 \mathrm{~m} \) is covered in 15 seconds.
We know that,
$Speed=\frac{Distance}{Time}$
Speed of the plane $=\frac{2196}{\frac{15}{60\times60}} \mathrm{~m} / \mathrm{hr}$
$=\frac{7905600}{15} \mathrm{~m} / \mathrm{hr}$
$=\frac{527040}{1000} \mathrm{~km} / \mathrm{hr}$
$=527.04 \mathrm{~km} / \mathrm{hr}$
Therefore, the speed of the aeroplane is $527.04 \mathrm{~km} / \mathrm{hr}$.
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