Question : A boat is moving away from an observation tower. It makes an angle of depression of 60° with an observer's eye when at a distance of 50 metres from the tower. After 8 seconds, the angle of depression becomes 30º. By assuming that it is running in still water, the approximate speed of the boat is:
Option 1: 33 km/h
Option 2: 42 km/h
Option 3: 45 km/h
Option 4: 50 km/h
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Correct Answer: 45 km/h
Solution : Given: The boat makes an angle of depression of 60º with an observer's eye when at a distance of 50 metres from the tower. After 8 seconds, the angle of depression becomes 30°. Solution: Let $CD = x$ m In $\triangle ABD$ $\tan 60°=\frac{AB}{AC}$ ⇒ $\sqrt{3}=\frac{AB}{50}$ ⇒ $AB=50\sqrt{3}$ In $\triangle ABD$ $\tan 30°=\frac{AB}{AC+CD}$ ⇒ $\frac{1}{\sqrt{3}}=\frac{50\sqrt{3}}{50+x}$ ⇒ $x=50\times 3 \ – \ 50$ ⇒ $x=100$ m Distance = 100 m, Time = 8 sec Speed = $\frac{\text{Distance}}{\text{Time}}$ Speed = $\frac{100\times 18}{8\times 5} = 45$ km/h. (multiplied by 18 and divided by 5 to get the speed in km/h) Hence, the correct answer is 45 km/h.
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