
Vector  B→B→ has magnitude 7.30 and direction 14.0° below the +*x*-axis. Vector  C→C→ has *x*-component *Cx* = −1.90 and *y*-component *Cy* = −6.70.
 

What angle does the resultant vector  C→+B→C→+B→  make with the +*x*-axis in clockwise direction? A positive angle is counterclockwise from the +*x*-axis.

To find the angle that the resultant vector \( \vec{R} = \vec{C} + \vec{B} \) makes with the \( +x \)-axis in the clockwise direction, follow these steps:

### Step 1: Determine the components of vector \( \vec{B} \)

Given:
- Magnitude of \( \vec{B} \): \( B = 7.30 \)
- Direction of \( \vec{B} \): \( 14.0^\circ \) below the \( +x \)-axis

Since \( 14.0^\circ \) is below the \( +x \)-axis, it’s equivalent to \( -14.0^\circ \) when measured counterclockwise from the \( +x \)-axis.

The components of \( \vec{B} \) are:
\[
B_x = B \cos(-14.0^\circ) \quad \text{and} \quad B_y = B \sin(-14.0^\circ)
\]

Substituting the values:
\[
B_x = 7.30 \cos(-14.0^\circ) \approx 7.08 \quad \text{(rounded to two decimal places)}
\]
\[
B_y = 7.30 \sin(-14.0^\circ) \approx -1.77 \quad \text{(rounded to two decimal places)}
\]

### Step 2: Combine the components of \( \vec{B} \) and \( \vec{C} \)

Given for vector \( \vec{C} \):
- \( C_x = -1.90 \)
- \( C_y = -6.70 \)

The components of the resultant vector \( \vec{R} = \vec{C} + \vec{B} \) are:
\[
R_x = C_x + B_x = -1.90 + 7.08 = 5.18
\]
\[
R_y = C_y + B_y = -6.70 + (-1.77) = -8.47
\]

### Step 3: Calculate the angle of the resultant vector

The angle \( \theta_R \) that the resultant vector makes with the \( +x \)-axis is given by:
\[
\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]
Substituting the values:
\[
\theta_R = \tan^{-1}\left(\frac{-8.47}{5.18}\right)
\]
\[
\theta_R \approx \tan^{-1}(-1.635) \approx -58.84^\circ
\]

### Step 4: Interpret the angle in the clockwise direction

The angle \( -58.84^\circ \) is measured counterclockwise from the \( +x \)-axis, which corresponds to \( 58.84^\circ \) clockwise. Therefore, the angle in the clockwise direction is:

\[
\theta_{\text{clockwise}} = 58.84^\circ
\]

So, the resultant vector \( \vec{C} + \vec{B} \) makes an angle of approximately \( 58.84^\circ \) clockwise from the \( +x \)-axis.

Would you like any further explanation or details? Here are some related questions you might find interesting:

1. How do you calculate the magnitude of the resultant vector \( \vec{R} \)?
2. Why is the direction below the \( +x \)-axis considered as a negative angle?
3. How would the situation change if the vectors were in three dimensions?
4. What are the trigonometric identities used in calculating vector components?
5. How would you approach this problem if the angle for \( \vec{B} \) were given above the \( +x \)-axis?

**Tip:** When dealing with angles and vector components, always ensure the angle is measured with respect to the correct axis, and consider the sign (positive or negative) of the components based on the vector's direction.

Learn how to calculate the angle that the resultant vector C+B makes with the +x-axis in the clockwise direction using vector components and trigonometric principles. This problem involves vectors with given magnitudes, components, and directions. Suitable for high school students studying vectors and trigonometry.

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