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We are given a problem where a ship travels along two bearings, and we need to calculate the final distance of the ship from its starting point. Here's how we can approach the problem:

### Known Information:
1. The ship first sails **18.5 nautical miles (NM)** on a bearing of **189°**.
2. The ship then turns and sails **47.8 NM** on a bearing of **317°**.

### Step-by-Step Approach:

1. **Represent the path of the ship as vectors:**
   - The two segments of the journey can be treated as vectors based on their distance and direction.
   
2. **Convert the bearings to standard Cartesian angles:**
   - Bearings are measured clockwise from the north, but we need to convert these into angles used in trigonometry (measured counterclockwise from the positive x-axis).
   - Bearing \(189^\circ\) corresponds to a trigonometric angle of \(189^\circ - 90^\circ = 99^\circ\).
   - Bearing \(317^\circ\) corresponds to a trigonometric angle of \(317^\circ - 90^\circ = 227^\circ\).

3. **Set up the coordinates:**
   - The position of the ship after each leg of the journey can be described using the cosine and sine functions to calculate the x and y components of the vector.

Let’s now calculate these using trigonometric components.

I'll perform the calculation for you.The distance of the ship from point \( A \) is approximately **39.22 nautical miles (NM)**.

Would you like further details or explanations on this solution? Here are five related questions for deeper understanding:

1. How can the use of vector components simplify solving navigation problems like this one?
2. Why do we convert bearings into trigonometric angles when solving such problems?
3. How would the problem change if the ship sailed a third leg on a different bearing?
4. What are the advantages of using trigonometric functions to break down direction and distance in navigation?
5. How could this problem be solved graphically using a scale and a compass?

**Tip:** When solving navigation problems with bearings, always double-check the conversion between bearings and standard trigonometric angles, as this step is crucial for accuracy.

Starting at point A, a ship sails 18.5 NM on a bearing of 189°, then turns and sails 47.8 NM on a bearing of 317°. Find the distance of the ship from point A.

Solve a navigation problem where a ship sails 18.5 NM at a bearing of 189° and then 47.8 NM at a bearing of 317°. Use vector components and trigonometric rules to find the final distance from the starting point. Suitable for Grades 10-12.

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