Question : Two ships are on the opposite of a lighthouse such that all three of them are collinear. The angles of depression of the two ships from the top of the lighthouse are 30° and 60°. If the ships are $230 \sqrt{3}$ m apart, then find the height of the lighthouse (in m).
Option 1: 175.4
Option 2: 165.2
Option 3: 172.5
Option 4: 180.5
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Correct Answer: 172.5
Solution :
Given: The distance between the ships is BC = $230\sqrt3$ m
Let the height of the lighthouse AD be $h$ m.
From $\triangle$ABD we get,
$\tan30°=\frac{AD}{BD}$ ⇒ BD = $h\sqrt3$
Again from $\triangle$ACD we get,
$\tan60°=\frac{AD}{CD}$ ⇒ CD = $\frac{h}{\sqrt3}$
According to the question,
$h\sqrt3+\frac{h}{\sqrt3}=230\sqrt3$
⇒ $3h+h=690$
⇒ $4h=690$
⇒ $h=\frac{690}{4}=172.5$ m
Hence, the correct answer is 172.5 m.
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