Two ships are sailing in the sea on the two sides of a lighthouse. The angle…
2024
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
- A.
173 m
- B.
200 m
- C.
273 m
- D.
300 m
Show answer & explanation
Correct answer: C
Angle-of-elevation problems use the tangent ratio, tan(angle) = opposite/adjacent, where the opposite side is the fixed vertical height and the adjacent side is the horizontal distance from its base. When two such horizontal distances lie on opposite sides of the same height, the total distance between the two far points equals the sum of the two individual horizontal distances.
Let AB be the lighthouse, with height 100 m, and let C and D be the positions of the two ships, with the angle of elevation ∠ACB = 30° and ∠ADB = 45°.
For the ship at C: tan 30° = AB/AC = 1/√3, so AC = AB × √3 = 100√3 m.
For the ship at D: tan 45° = AB/AD = 1, so AD = AB = 100 m.
The distance between the ships is CD = AC + AD = 100√3 + 100 = 100(√3 + 1) m.
Using √3 ≈ 1.73, CD ≈ 100 × 2.73 = 273 m.
Cross-check: tan 30° ≈ 0.577 gives AC ≈ 173 m, and tan 45° = 1 gives AD = 100 m; their sum, 273 m, matches the exact expression 100(√3 + 1) m derived above, confirming the result independently.
