Question : An aeroplane flying horizontally at a height of 3 km above the ground is observed at a certain point on earth to subtend an angle of $60^\circ$. After 15 seconds of flight, its angle of elevation is changed to $30^\circ$. The speed of the aeroplane (Take $\sqrt{3}=1.732$) is:
Option 1: 230.63 m/s
Option 2: 230.93 m/s
Option 3: 235.85 m/s
Option 4: 236.25 m/s
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Correct Answer: 230.93 m/s
Solution : AB = CD = 3 km = 3000 m In $\triangle$AOB, $\tan60^{\circ}=\frac{AB}{OB}$ ⇒ $OB=\frac{3000}{\sqrt3}=1000\sqrt3$ m In $\triangle$COD, $\tan30^{\circ}=\frac{CD}{OC}$ ⇒ $OC=3000\sqrt3$ m So, BC = AD = ($3000\sqrt3–1000\sqrt3)=2000\sqrt3$ m Therefore, the speed of the aeroplane = $\frac{2000\sqrt3}{15}$ m/s = 230.93 m/s Hence, the correct answer is 230.93 m/s.
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