Question : 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: (take $\sqrt{3}= 1.73$)
Option 1: 173 metres
Option 2: 200 metres
Option 3: 273 metres
Option 4: 300 metres
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Correct Answer: 273 metres
Solution : Let, BD be the lighthouse and A and C be the positions of the ships. Then, BD=100 m, $\angle$BAD=30°, $\angle$BCD=45° In $\triangle$ABD, we have $\tan 30° = \frac{BD}{BA}$ ⇒ $\frac{1}{\sqrt{3}}$ = $\frac{100}{BA}$ ⇒ BA = 100$\sqrt{3}$ In $\triangle$CBD, we have $ \tan 45° = \frac{BD}{BC}$ ⇒ 1 = $\frac{100}{BC}$ ⇒ BC = 100 m Distance between the two ships = AC = BA + BC = 100$\sqrt{3}$ + 100 = 100($\sqrt{3}$ + 1) = 100(1.73 + 1) = 100 × 2.73 = 273 m Hence, the correct answer is 273 m.
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