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 it is at a distance of 100 m from the tower. After 10 seconds the angle of depression becomes 30°. What is the approximate speed of the boat, assuming it is running in still water?
Answer: D
At \( 45^{\circ} \) the horizontal distance equals the height, so the tower measured to the man's eye is 100 m.
After 10 seconds the depression is \( 30^{\circ} \):
\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = 100\sqrt{3} = 173.2 \text{ m} \]
\[ \text{distance covered} = 173.2 - 100 = 73.2 \text{ m} \]
\[ \text{speed} = \frac{73.2}{10} = 7.32 \text{ m/s} = 7.32 \times \frac{18}{5} = 26.35 \approx 26.28 \text{ km/h} \]
Only the difference of the two distances is travelled in those 10 seconds; using 173.2 m outright would assume the boat set off from the tower.
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