Two ships are on the opposite sides 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 $30^\circ$ and $60^\circ$. If the ships are $230\sqrt{3} ext{ m}$ apart, then find the height of the lighthouse (in m).

Question

Two ships are on the opposite sides 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 $30^\circ$ and $60^\circ$. If the ships are $230\sqrt{3} 	ext{ m}$ apart, then find the height of the lighthouse (in m).

Accepted Answer

**Solution:**

1. Let the height of the lighthouse be $h$ meters.

2. The two ships are on opposite sides of the lighthouse, and all three points are collinear.

3. Let the distances of the ships from the base of the lighthouse be $BD$ and $CD$ respectively.

4. From the angle of depression $30^\circ$ to ship at $B$:
   $$
   \tan 30^\circ = \frac{AD}{BD} = \frac{h}{BD} \implies BD = \frac{h}{\tan 30^\circ} = \frac{h}{\frac{1}{\sqrt{3}}} = h \sqrt{3}
   $$

5. From the angle of depression $60^\circ$ to ship at $C$:
   $$
   \tan 60^\circ = \frac{AD}{CD} = \frac{h}{CD} \implies CD = \frac{h}{\tan 60^\circ} = \frac{h}{\sqrt{3}}
   $$

6. The total distance between the ships is given as:
   $$
   BC = BD + CD = h \sqrt{3} + \frac{h}{\sqrt{3}} = 230 \sqrt{3}
   $$

7. Simplify the left side:
   $$
   h \sqrt{3} + \frac{h}{\sqrt{3}} = h \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = h \left( \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) = h \times \frac{4}{\sqrt{3}}
   $$

8. So,
   $$
   h \times \frac{4}{\sqrt{3}} = 230 \sqrt{3} \implies h = \frac{230 \sqrt{3} \times \sqrt{3}}{4} = \frac{230 \times 3}{4} = \frac{690}{4} = 172.5
   $$

**Answer:** $172.5$ meters

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