Question : The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds of flight, the elevation changes to 30°. If the aeroplane is flying at a height of $1500\sqrt{3}$ metre, find the speed of the plane:
Option 1: 300 m/sec
Option 2: 200 m/sec
Option 3: 100 m/sec
Option 4: 150 m/sec
Correct Answer: 200 m/sec
Solution :
Height of the aeroplane = BD = CE = $1500\sqrt{3}$ m and $\angle$ BAE = 60°, $\angle$ CAE = 30°
In $\Delta$ ADB,
$\tan 60°=\frac{1500\sqrt{3}}{AD}$
⇒ $\sqrt{3}=\frac{1500\sqrt{3}}{AD}$
⇒ AD = $\frac{1500\sqrt{3}}{\sqrt{3}}$
⇒ AD = 1500 m
In $\Delta$ CAE,
$\tan 30°=\frac{1500\sqrt{3}}{AE}$
⇒ $\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AE}$
⇒ AE = $1500\sqrt{3}×\sqrt{3}$
⇒ AE = 4500 m
Distance covered by plane in 15 seconds:
BC = DE = AE – AD = 4500 – 1500 = 3000 m
So, speed of plane = $\frac{3000}{15}$ = 200 m/s
Hence, the correct answer is 200 m/sec.
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