A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of $$45^\circ$$ with the man's eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression become $$30^\circ$$. What is the approximate speed of the boat, assuming that it is running in still water?

Let AB be the tower and BC = 60 m. It takes 5 seconds for the boat to go from C to point D.

In right $$\triangle$$ ABC,

=> $$tan(45^\circ)=\frac{AB}{BC}$$

=> $$1=\frac{AB}{60}$$

=> $$AB=60$$ m ----------------(i)

Similarly, in right $$\triangle$$ ABD,

=> $$tan(30^\circ)=\frac{AB}{BD}$$

=> $$\frac{1}{\sqrt3}=\frac{60}{BD}$$

=> $$BD=60\sqrt3$$ m ----------------(ii)

Thus, CD = $$60\sqrt3-60=60(\sqrt3-1)$$ m

Using speed = distance/time, => Speed of boat = $$\frac{60(\sqrt3-1)}{5}$$

= $$12(\sqrt3-1)=12\times0.732=8.784$$ m/s

$$\therefore$$ Speed of boat (in km/hr) = $$8.784\times3.6\approx32$$ km/hr

=> Ans - (A)