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Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is :_________?

Mathematicsmedium
A173 m
B200 m
273 m
D300 m Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ∠ACB = 300 and ∠ADB = 45°. AB/AC = tan 30° = 1/√3 AC = AB X √3 = 100√3 m. AB/AD = tan 45° = 1 ⇒ AD = AB = 100 m. CD = (AC + AD) = (100√3 + 100) m = 100 (√3 +1) m = (100 X 2.73) m = 273 m.

Correct Answer: C. 273 m

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