Instructions

Read the following information carefully and answer the questions

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°.

Question 110

If both are moving in same direction and the submarine is ahead of the warship in both the situations, then the speed of the warship, if the ratio of the speed of warship to that of the submarine is 2 : 1, is:

Solution

Let the initial position of the warship be A and initial and final positions of the submarine be P and Q respectively.

$$AP = 50 km$$

Ratio of speed of warship to submarine = $$2 : 1$$

=> $$AB = 2 PQ$$ ---------------Eqn(I)

In $$\triangle$$ APN

=> $$sin 30 = \frac{AN}{AP}$$

=> $$\frac{1}{2} = \frac{AN}{50}$$

=> $$AN = \frac{50}{2} = 25 km$$

=> $$BM = AN = 25 km$$

Similarly, $$PN = 25\sqrt{3} = AR$$

$$\therefore$$ In $$\triangle$$ BMQ

=> $$sin 60 = \frac{BM}{BQ}$$

=> $$\frac{\sqrt{3}}{2} = \frac{25}{BQ}$$

=> $$BQ = \frac{50}{\sqrt{3}} km$$

Similarly,  $$MQ = \frac{25}{\sqrt{3}}$$

Let $$RB = PM = x$$ and using Eqn (I), we get :

=> $$(AR + RB) = 2 (PM + MQ)$$

=> $$(25\sqrt{3} + x) = 2 (x + \frac{25}{\sqrt{3}}$$

=> $$(25\sqrt{3} + x) = 2x + \frac{50}{\sqrt{3}}$$

=> $$2x - x = 25\sqrt{3} - \frac{50}{\sqrt{3}}$$

=> $$x = \frac{25}{\sqrt{3}}$$

$$\therefore AB = 25\sqrt{3} + \frac{25}{\sqrt{3}}$$

=> $$AB = \frac{100}{\sqrt{3}}$$

$$\therefore$$ Speed of warship = $$2 \times \frac{100}{\sqrt{3}} = \frac{200}{\sqrt{3}} km/h$$


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