The angle of elevation of an aeroplane from point A on the ground is $60^{o}$. After flight of 15 seconds, the angle of elevation changes to $30^{o}$. If the aeroplane is flying at a constant height of $1500\sqrt{3}\ m$, find the speed of the plane in km/hr.
Given: The angle of elevation of an aeroplane from point A on the ground is $60^{o}$. After flight of 15 seconds, the angle of elevation changes to $30^{o}$, height of aeroplane$=1500\sqrt{3}\ m$.
To do: To find the speed of the plane in $km/hr$.
Solution:
Let BC be the height at watch the aeroplane is observed from point A.
Then, $BC = 1500\sqrt{3}\ m$
In 15 seconds, the aeroplane moves from point A to D.
A and D are the points where the angles of elevation $60^{o}$ and $30^{o}$
are formed respectively.
Let $BA=x$ meter and $AD = y$ meter
$BC = x + y$
In $\vartriangle CBA,$
$tan60^{o}=\frac{BC}{BA}$
$\sqrt{3} =\frac{1500\sqrt{3}}{x}$
$\Rightarrow x=1500\ m\ \ \ .............\left( 1\right)$
In $\vartriangle CBD$,
$tan30^{o}=\frac{BC}{BD}$
$\frac{1}{\sqrt{3}} =\frac{1500\sqrt{3}}{x+y}$
$\Rightarrow x+y=1500\times 3=4500$
$\Rightarrow 1500+y=4500$
$\Rightarrow y=4500-1500$
$\Rightarrow y=3000\ m\ \ ..............\left( 2\right)$
We know that, the aeroplane moves from point A to D in 15 seconds and the distance covered is $3000\ m$.
Speed$=\frac{Distance}{Time}$
Speed$=\frac{3000}{15}$
$\Rightarrow$ Speed$=200\ m/s$
Converting it to $km/h$
Speed$=200\times \frac{18}{5}$
$=720\ km/h$
Thus, The speed of the aeroplane is $720\ km/h$.
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