Question

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships as \(30^\circ\) and \(45^\circ\) respectively. If the lighthouse is \(100\) m high, the distance between the two ships is:

• A. \(173\ \text{m}\)
• B. \(200\ \text{m}\)
• C. \(273\ \text{m}\)
• D. \(300\ \text{m}\)

Hint:

For opposite-side angle of elevation problems, compute horizontal distances separately using \(\tan \theta\) and add them.

Answer 1

The Correct Option is C

Step 1 (Understanding the problem).
Let the base of the lighthouse be \(O\), the top of the lighthouse be \(T\).
Let ship A be on one side at distance \(d_1\) from \(O\) with angle of elevation \(30^\circ\).
Let ship B be on the other side at distance \(d_2\) from \(O\) with angle of elevation \(45^\circ\).
Height of the lighthouse \(OT = 100\) m.
Step 2 (Using \(\tan \theta\) for each ship).
For ship A: \[ \tan 30^\circ = \frac{OT}{d_1} \frac{1}{\sqrt{3}} = \frac{100}{d_1} \] \[ d_1 = 100 \sqrt{3} \ \text{m} \] For ship B: \[ \tan 45^\circ = \frac{OT}{d_2} 1 = \frac{100}{d_2} \] \[ d_2 = 100 \ \text{m} \] Step 3 (Total distance between the ships).
Since the ships are on opposite sides of the lighthouse: \[ \text{Distance} = d_1 + d_2 = 100\sqrt{3} + 100 \] Using \(\sqrt{3} \approx 1.732\): \[ \text{Distance} \approx 100(1.73(b) + 100 = 173.2 + 100 = 273.2 \ \text{m} \] Step 4 (Conclusion).
The distance is approximately \(273\) m, which matches Option 3.
\[ \boxed{273\ \text{m (Option (c)}} \]