This is a bearings and distances problem. We need to find the bearing of point $C$ from point $A$.

Given:

• A man starts at point $A$.
• Walks 2 km on a bearing of $017^\circ$ (from the north).
• Then walks 3 km on a bearing of $107^\circ$ to reach point $C$.

To solve:

We will treat the journey as a vector problem and use trigonometry (sine and cosine rules) to determine the bearing of $C$ from $A$.

1. Step 1: Plot the vectors.

   • Vector $AB$ represents the first leg of the journey, 2 km at a bearing of $017^\circ$.
   • Vector $BC$ represents the second leg, 3 km at a bearing of $107^\circ$.

2. Step 2: Break the vectors into components.

   • For $AB$: Bearing is $017^\circ$, so: $x_1 = 2 \sin(17^\circ), \quad y_1 = 2 \cos(17^\circ)$
   • For $BC$: Bearing is $107^\circ$, so: $x_2 = 3 \sin(107^\circ), \quad y_2 = 3 \cos(107^\circ)$ These components will help us calculate the final position of $C$ relative to $A$.

3. Step 3: Combine the components.

   • The total horizontal (x) and vertical (y) components will give us the overall displacement from $A$ to $C$.

4. Step 4: Calculate the bearing. The bearing from $A$ to $C$ can be found using: $\theta = \tan^{-1} \left( \frac{\text{total x}}{\text{total y}} \right)$ Adjust the angle to ensure it's measured from north in a clockwise direction.

I'll calculate these values next.The bearing of point $C$ from point $A$ is approximately $73.31^\circ$.

Would you like further details or clarification on any part of this process?

Here are 5 related questions for further practice:

1. How do you convert bearings from degrees to radians?
2. What is the difference between true bearing and compass bearing?
3. How would the bearing change if the man walked south instead of north first?
4. How can you use the cosine rule to solve for bearings in triangle problems?
5. How do you apply vector addition in solving navigation problems like this?

Tip: When calculating bearings, always measure clockwise from north, and ensure the final bearing is adjusted to lie between 0° and 360°.