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The angle of elevation of an aeroplane from a point A on the ground is $60^{\circ}$ . After a flight of $30$ seconds, the angle of elevation changes to $30^{\circ}$ . If the plane is flying at a constant height of $3600\sqrt{3}\; metres,$ metres, find the speed of the aeroplane.

Answers (1)

Given, $\angle BAE =60^{\circ}$

$\angle CAD=30^{\circ}$

$BE=CD=3600\sqrt{3}$ meters

time = $30 \; seconds$

To find - speed of plane

$speed =\frac{distance}{time}$

$\Rightarrow distance=BC$ ; we need to find BC

Now in $\Delta ABE$

$\tan 60^{\circ}=\frac{BE}{AE}$

$\sqrt{3}=\frac{3600\sqrt{3}}{AE}\Rightarrow AE=3600\; m$

Now in $\Delta ACD$

$\tan 30^{\circ}=\frac{CD}{AD}$

$\frac{1}{\sqrt{3}}=\frac{3600\sqrt{3}}{AD}\Rightarrow AD=3600\times 3$

$\Rightarrow AD=10800$

$\Rightarrow AD=AE+ED$

$\Rightarrow ED=AD-AE$

$\Rightarrow BC=AD-AE=10800-3600\; \; \; \; \; \; (\therefore ED=BC)$

$\Rightarrow BC=7200\; Meters$

$\Rightarrow$ Speed of aeroplane = $=\frac{7200}{30}=240\; m/s$

speed of aeroplane $=240\; m/s$

Posted by

Safeer PP

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