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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