Question : A helicopter, at an altitude of 1500 metres, finds that two ships are sailing towards it, in the same direction. The angles of depression of the ships as observed from the helicopter are $60^{\circ}$ and $30^{\circ}$, respectively. Distance between the two ships, in metres, is:
Option 1: $1000\sqrt{3}$
Option 2: $\frac{1000}{{\sqrt3}}$
Option 3: $500{\sqrt3}$
Option 4: $\frac{500}{{\sqrt3}}$
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Correct Answer: $1000\sqrt{3}$
Solution : Let D be the aeroplane. A and B are two ships. Given: $CD = 1500 \text{ m,} \angle EDA = 30^{\circ}, \angle EDB = 60^{\circ}$ In $\triangle DBC,$ $\tan 60^{\circ} = \frac{DC}{BC}$ $\Rightarrow \sqrt{3} = \frac{1500}{BC}$ $\Rightarrow BC = \frac{1500}{\sqrt{3}} \text{ m}$ In $\triangle ADC, $ $\tan 30^{\circ} = \frac{DC}{AC}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{AB + BC}$ $\Rightarrow AB + BC = 1500\sqrt{3}$ $\Rightarrow AB = 1500\sqrt{3} - \frac{1500}{\sqrt{3}}$ $\Rightarrow AB = 1500(\sqrt{3} - \frac{1}{\sqrt{3}})$ $\Rightarrow AB = 1500( \frac{3-1}{\sqrt{3}})$ $\Rightarrow AB = \frac{3000}{\sqrt{3}}$ $\Rightarrow AB = 1000\sqrt{3}$ metres. Hence, the correct answer is $1000\sqrt{3}$ metres.
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