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A warship and a submarine (completely submerged in water) are moving horizontally in a straight line. The Captain of the warship observers that the submarine makes an angle of depression of 30°, and the distance between them from the point of observation is 50 km. After 30 minutes, the angle of depression becomes 60°.
Let the initial position of the warship be A and initial and final positions of the submarine be P and Q respectively.
$$AP = 50 km$$
In $$\triangle$$ APN
=> $$sin 30 = \frac{AN}{AP}$$
=> $$\frac{1}{2} = \frac{AN}{50}$$
=> $$AN = \frac{50}{2} = 25 km$$
=> $$BM = AN = 50 km$$
$$\therefore$$ In $$\triangle$$ BMQ
=> $$sin 60 = \frac{BM}{BQ}$$
=> $$\frac{\sqrt{3}}{2} = \frac{25}{BQ}$$
=> $$BQ = \frac{50}{\sqrt{3}} km$$
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