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
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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