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The image you provided shows a vector addition problem where you need to find the magnitudes of two velocities, \( v_A \) and \( v_B \), given that their resultant \( v_{\text{tot}} = v_A + v_B \), forms specific angles with respect to the x and y axes.

Here’s how to approach the problem:

### Given:
- \( \theta = 23.0^\circ \) (the angle between the resultant vector \( v_{\text{tot}} \) and the y-axis),
- \( \phi = 26.5^\circ \) (the angle between \( v_A \) and the x-axis),
- \( v_{\text{tot}} = 5.56 \, \text{m/s} \) (magnitude of the resultant velocity).

### Step 1: Resolve the components of the velocities
We can resolve the vectors using trigonometry. From the triangle:

1. \( v_A \) has a known angle \( 26.5^\circ \) with the x-axis, so:
   - \( v_A \cos(26.5^\circ) \) is the x-component of \( v_A \),
   - \( v_A \sin(26.5^\circ) \) is the y-component of \( v_A \).

2. \( v_{\text{tot}} \) is the resultant vector, and its components are:
   - \( v_{\text{tot}} \sin(23.0^\circ) \) is the x-component of \( v_{\text{tot}} \),
   - \( v_{\text{tot}} \cos(23.0^\circ) \) is the y-component of \( v_{\text{tot}} \).

### Step 2: Set up the equations
The total velocity \( v_{\text{tot}} \) is the sum of the components of \( v_A \) and \( v_B \). Thus, we can write the equations for the x and y components as:

1. **For the x-component:**
   \[
   v_A \cos(26.5^\circ) = v_{\text{tot}} \sin(23.0^\circ)
   \]

2. **For the y-component:**
   \[
   v_A \sin(26.5^\circ) + v_B = v_{\text{tot}} \cos(23.0^\circ)
   \]

### Step 3: Solve for \( v_A \) and \( v_B \)
We can solve these two equations to find the magnitudes of \( v_A \) and \( v_B \).

1. From the first equation:
   \[
   v_A = \frac{v_{\text{tot}} \sin(23.0^\circ)}{\cos(26.5^\circ)}
   \]

   Plugging in the values:
   \[
   v_A = \frac{5.56 \times \sin(23.0^\circ)}{\cos(26.5^\circ)}
   \]
   \[
   v_A \approx 2.27 \, \text{m/s}
   \]

2. Now, substitute \( v_A \) into the second equation to find \( v_B \):
   \[
   v_A \sin(26.5^\circ) + v_B = v_{\text{tot}} \cos(23.0^\circ)
   \]
   \[
   2.27 \times \sin(26.5^\circ) + v_B = 5.56 \times \cos(23.0^\circ)
   \]
   Solving for \( v_B \):
   \[
   v_B = 5.56 \times \cos(23.0^\circ) - 2.27 \times \sin(26.5^\circ)
   \]
   \[
   v_B \approx 4.99 \, \text{m/s}
   \]

### Final Answer:
- \( v_A \approx 2.27 \, \text{m/s} \)
- \( v_B \approx 4.99 \, \text{m/s} \)

Would you like further clarification or more details on this solution?

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Here are 5 related questions to expand on this concept:
1. How do you decompose any vector into its x and y components?
2. What would happen to the solution if the angle between \( v_A \) and the x-axis changed?
3. How can you use the law of cosines to solve similar vector problems?
4. What is the geometric interpretation of vector addition?
5. How would you apply this method to find the relative velocities in a moving frame?

### Tip: Always double-check the units and angles when working with trigonometric identities in physics problems to avoid calculation errors!

Find the magnitudes of velocities v_A and v_B in the figure below, where θ = 23.0° and v_tot = 5.56 m/s.

Learn how to calculate the magnitudes of velocities v_A and v_B using vector addition and trigonometry. This problem involves resolving components of vectors with angles θ = 23.0° and φ = 26.5°, along with the given resultant velocity v_tot = 5.56 m/s. Suitable for high school physics students.

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