Discussion
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° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
Answer: B
At \( 45^{\circ} \) the horizontal distance equals the height, so the tower (measured to the man's eye) is \( 100 \) m.
Ten seconds later the depression is \( 30^{\circ} \):
\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = 100\sqrt{3} = 173.2 \text{ m} \]
\[ \text{distance covered} = 173.2 - 100 = 73.2 \text{ m in } 10 \text{ s} \]
\[ \text{speed} = 7.32 \text{ m/s} = 7.32 \times \frac{18}{5} = 26.35 \text{ km/h} \approx 26.28 \text{ km/h} \]
Use the difference of the two distances, not \( 173.2 \) itself - that would give the speed as if the boat had started at the tower.
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