In the sum A + B = C, vector A has a magnitude of 12.0 m and is angled 40° counterclockwise from 
the +x direction, and vector C has magnitude of 15.0 m and is angled 20° counterclockwise from 
the −x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B?

To solve for the magnitude and direction of vector **B** in the equation **A + B = C**, we need to use vector components and trigonometry. Here's the step-by-step process:

### Given:
- Magnitude of **A**: \( |A| = 12.0 \, \text{m} \)
- Direction of **A**: \( 40^\circ \) counterclockwise from the \( +x \)-axis.
- Magnitude of **C**: \( |C| = 15.0 \, \text{m} \)
- Direction of **C**: \( 20^\circ \) counterclockwise from the \( -x \)-axis (this is equivalent to \( 180^\circ - 20^\circ = 160^\circ \) relative to the \( +x \)-axis).

### Step 1: Find the components of vectors A and C.
We can break down vectors **A** and **C** into their x and y components using trigonometry.

For vector **A**:
- \( A_x = |A| \cdot \cos(40^\circ) = 12.0 \cdot \cos(40^\circ) \)
- \( A_y = |A| \cdot \sin(40^\circ) = 12.0 \cdot \sin(40^\circ) \)

For vector **C**:
- \( C_x = |C| \cdot \cos(160^\circ) = 15.0 \cdot \cos(160^\circ) \)
- \( C_y = |C| \cdot \sin(160^\circ) = 15.0 \cdot \sin(160^\circ) \)

### Step 2: Calculate components of vector **B**.
Since \( \mathbf{A} + \mathbf{B} = \mathbf{C} \), we have:
- \( B_x = C_x - A_x \)
- \( B_y = C_y - A_y \)

### Step 3: Calculate the magnitude and direction of vector **B**.
- Magnitude of **B**: 
  \[
  |B| = \sqrt{B_x^2 + B_y^2}
  \]
- Angle of **B** relative to the \( +x \)-axis: 
  \[
  \theta_B = \tan^{-1} \left( \frac{B_y}{B_x} \right)
  \]

Let's compute these values.import math

# Given values
A_magnitude = 12.0  # magnitude of A in meters
A_angle = 40  # angle of A in degrees relative to +x axis
C_magnitude = 15.0  # magnitude of C in meters
C_angle = 160  # angle of C in degrees relative to +x axis

# Convert angles to radians for trigonometric calculations
A_angle_rad = math.radians(A_angle)
C_angle_rad = math.radians(C_angle)

# Components of vector A
A_x = A_magnitude * math.cos(A_angle_rad)
A_y = A_magnitude * math.sin(A_angle_rad)

# Components of vector C
C_x = C_magnitude * math.cos(C_angle_rad)
C_y = C_magnitude * math.sin(C_angle_rad)

# Components of vector B (B_x = C_x - A_x, B_y = C_y - A_y)
B_x = C_x - A_x
B_y = C_y - A_y

# Magnitude of vector B
B_magnitude = math.sqrt(B_x**2 + B_y**2)

# Angle of vector B relative to +x axis (in degrees)
B_angle = math.degrees(math.atan2(B_y, B_x))

B_magnitude, B_angleThe magnitude and angle of vector **B** are as follows:

- **Magnitude of B**: \( 23.43 \, \text{m} \)
- **Angle of B** relative to the \( +x \)-axis: \( -173.67^\circ \) (which means it is slightly below the negative \( x \)-axis).

Would you like more details or clarifications?

Here are some related questions for further practice:
1. How would the magnitude of vector **B** change if vector **C** was doubled?
2. What happens to the angle of vector **B** if the angle of **C** is reduced by 30 degrees?
3. How would you approach this problem if vector **A** was vertical instead of angled?
4. What is the direction of vector **A + B** if vector **C** was rotated 90 degrees?
5. How do you convert between Cartesian and polar coordinates in vector problems?

**Tip**: Always check the quadrant in which a vector lies when calculating the angle to avoid sign errors in trigonometric functions.

Learn how to calculate vector B in the equation A + B = C using trigonometry. Given vectors A and C with their magnitudes and directions, find the magnitude and angle of vector B relative to the +x-axis. Suitable for advanced high school students studying vectors and trigonometry.

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