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To solve this problem, we can break down the boat's journey into vectors, using trigonometry to determine the resultant vector that represents the direct path back to the starting point.

### Given Information:
1. The boat travels 8 km on a bearing of \(315^\circ\).
2. Then, it travels 6 km on a bearing of \(045^\circ\).

### Step 1: Represent Each Journey as Vectors
We can represent each leg of the journey in terms of eastward (x-axis) and northward (y-axis) components.

1. **For the first journey (8 km at \(315^\circ\)):**
   - A bearing of \(315^\circ\) is equivalent to moving 45 degrees below the westward line.
   - Thus, we have:
     \[
     x_1 = 8 \cos(315^\circ) = 8 \times \frac{\sqrt{2}}{2} = 8 \times 0.7071 = 5.656
     \]
     \[
     y_1 = 8 \sin(315^\circ) = 8 \times \frac{-\sqrt{2}}{2} = -5.656
     \]

   - Therefore, the first vector is approximately \((5.656, -5.656)\).

2. **For the second journey (6 km at \(045^\circ\)):**
   - A bearing of \(045^\circ\) corresponds to 45 degrees above the eastward line.
   - We get:
     \[
     x_2 = 6 \cos(45^\circ) = 6 \times \frac{\sqrt{2}}{2} = 6 \times 0.7071 = 4.242
     \]
     \[
     y_2 = 6 \sin(45^\circ) = 6 \times \frac{\sqrt{2}}{2} = 4.242
     \]

   - Therefore, the second vector is approximately \((4.242, 4.242)\).

### Step 2: Find the Resultant Displacement
Now, add the two vectors to get the total displacement:
\[
x = x_1 + x_2 = 5.656 + 4.242 = 9.898
\]
\[
y = y_1 + y_2 = -5.656 + 4.242 = -1.414
\]

### Step 3: Calculate the Bearing for the Return Journey
To find the bearing back to the starting point, we need the angle of the resultant vector with respect to the north.

1. **Calculate the angle**:
   \[
   \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-1.414}{9.898}\right)
   \]
   Calculating this angle:
   \[
   \theta \approx -8.13^\circ
   \]

2. **Convert to Bearing**:
   Since bearings are measured clockwise from north, the bearing will be:
   \[
   360^\circ - 8.13^\circ \approx 351.87^\circ
   \]

### Answer:
The bearing on which the boat should travel to return directly to the starting point is approximately **352°**.

Would you like further details on any part of this solution?

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Here are five related questions to expand your understanding:

1. How do you calculate bearings when moving in different directions on a map?
2. How can vector components help in understanding displacement?
3. What trigonometric functions are most commonly used in navigation?
4. Why is understanding angles and bearings important in navigation?
5. How does converting angles to bearings differ from regular angle measurement?

**Tip:** When dealing with bearings, remember they always start from the north (0°) and go clockwise up to 360°.

A boat travels 8 km on a bearing of 315° and then a further 6 km on a bearing of 045°. Find the bearing on which the boat should travel to return directly to the starting point.

Solve a navigation problem involving a boat's return bearing after traveling on two bearings. This solution covers trigonometry, vector components, and angle calculation. Ideal for students in Grades 9-12.

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