- A100(√3 − 1) m
- B100(√3 + 1) m
- C100√3 m
- D100 m
Step-by-Step Solution
Distances: 100 cot 30° = 100√3 and 100 cot 45° = 100. The ship sailed 100(√3 − 1) m ≈ 73.2 m.
Answer: A) 100(√3 − 1) m
From the top of a 100 m lighthouse, the angle of depression of a ship sailing straight towards it changes from 30° to 45°. How far did the ship sail in that time?
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