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Let's solve this! The problem involves bearings and the displacement of a ship. Here's how we can approach it step-by-step.

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### **Problem Recap and Setup:**
- **Initial Bearing:** N 20° E (this means the ship is located at a 20° angle east of due north from the lighthouse).
- **Travel:** The ship sails **5.6 miles south**.
- **New Bearing:** After traveling, the new bearing is now N 30° E.
- **Goal:** Find the **distance between the ship and the lighthouse** at both the **initial and final positions**.

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### **Solution Outline:**

We can represent this scenario using **trigonometry**. Let's assume the lighthouse is at the origin \((0, 0)\) and express the positions of the ship initially and finally in coordinates.

1. **Setup of Coordinates:**
   - Let \(d\) be the initial distance between the ship and the lighthouse (this is what we are solving for).
   - Using the bearing N 20° E:
     - \(x_1 = d \sin(20^\circ)\)
     - \(y_1 = d \cos(20^\circ)\)

   This gives the **initial position** of the ship as \((x_1, y_1) = \left(d \sin(20^\circ), d \cos(20^\circ)\right)\).

2. **After Moving South:**
   - Moving **5.6 miles south** means subtracting 5.6 from the y-coordinate:
     \[ y_2 = d \cos(20^\circ) - 5.6 \]
   - The x-coordinate remains the same: \(x_2 = d \sin(20^\circ)\).

3. **New Bearing Condition (N 30° E):**
   - Now, the **new bearing** is N 30° E. This gives us:
     \[ \tan(30^\circ) = \frac{x_2}{y_2} = \frac{d \sin(20^\circ)}{d \cos(20^\circ) - 5.6} \]

4. **Solve for \(d\):**
   Substituting values for trigonometric functions:
   - \(\sin(20^\circ) \approx 0.3420\)
   - \(\cos(20^\circ) \approx 0.9397\)
   - \(\tan(30^\circ) = \frac{\sqrt{3}}{3} \approx 0.5774\)

   Now plug these into the equation:
   \[
   0.5774 = \frac{0.3420 \cdot d}{0.9397 \cdot d - 5.6}
   \]

5. **Simplify and Solve:**
   Multiply both sides by \((0.9397 \cdot d - 5.6)\):
   \[
   0.5774 \cdot (0.9397 \cdot d - 5.6) = 0.3420 \cdot d
   \]
   Distribute on the left:
   \[
   0.5426 \cdot d - 3.236 = 0.3420 \cdot d
   \]
   Move terms involving \(d\) to one side:
   \[
   0.5426 \cdot d - 0.3420 \cdot d = 3.236
   \]
   \[
   0.2006 \cdot d = 3.236
   \]
   Solve for \(d\):
   \[
   d = \frac{3.236}{0.2006} \approx 16.1
   \]

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### **Final Answer:**
The initial distance between the ship and the lighthouse is approximately **16.1 miles**.

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Do you need further clarifications or any extra steps?

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### **5 Follow-up Questions:**
1. How do we interpret the N 20° E bearing in terms of navigation?
2. Can you use the Law of Cosines to verify this solution?
3. What is the significance of subtracting the southward displacement from the y-coordinate?
4. How would the solution change if the ship moved north instead of south?
5. How are bearings different from standard angle measurements?

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### **Tip:**  
Always make sure to visualize bearings as angles measured clockwise from the north to avoid confusion.

The bearing of a ship from a lighthouse was found to be N 20° E. After the ship sailed 5.6 miles due south, the new bearing was N 30° E. Find the distance between the ship and the lighthouse at each location.

Learn how to calculate the distance between a ship and a lighthouse based on bearing changes from N 20° E to N 30° E after a 5.6-mile southward sail. This problem involves trigonometric formulas and bearings suitable for high school students.

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