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This is a navigational problem involving bearings and distances. Let’s break down the given information:

1. A ship leaves **Shanghai** on a bearing of **030°** (30° clockwise from north) and sails **500 km** to **Jeju**.
2. At **Jeju**, the ship changes its direction to a bearing of **100°** and sails a distance of **320 km** to reach **Nagasaki**.
3. The task is to calculate the bearing and distance for the ship to return from **Nagasaki** to **Shanghai**.

To solve this, we can:
- Treat the paths as vectors and use trigonometry (Law of Cosines and Law of Sines) to determine the direct distance from Nagasaki to Shanghai.
- Then, calculate the bearing for the ship to sail back.

Let's start solving this step-by-step.

### Step 1: Represent the Paths as Vectors
We need to break down each segment into components based on the given bearings and distances:
- From Shanghai to Jeju (bearing = 030°, distance = 500 km),
- From Jeju to Nagasaki (bearing = 100°, distance = 320 km).

### Step 2: Calculate the Distance from Nagasaki to Shanghai
To find this, we can use the Law of Cosines, treating the triangle formed by Shanghai, Jeju, and Nagasaki as a triangle with known side lengths and angles.

I'll calculate this and provide the detailed solution.

A ship leaves Shanghai on a bearing of 030°. It sails a distance of 500 km to Jeju. In Jeju, the ship changes direction to a bearing of 100°. It sails a distance of 320 km to reach Nagasaki. Calculate the bearing and distance for the ship to return from Nagasaki to Shanghai.

Solve a navigational problem using trigonometry to calculate the bearing and distance for a ship to return from Nagasaki to Shanghai. Key concepts include vectors, bearings, and the laws of cosines and sines. Suitable for Grades 10-12.

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