https://file.aiquickdraw.com/mathbot/file/1726199398998xgtytn8o.jpg

To solve this problem, we need to calculate the magnitude and direction of the vector \(-\mathbf{A} - \mathbf{B}\) given the vectors \(\mathbf{A}\) and \(\mathbf{B}\) in the figure.

### Step 1: Extract Vector Information
From the image, we have:
- **Vector \(\mathbf{A}\):** 8.0 m at 30° above the negative x-axis.
- **Vector \(\mathbf{B}\):** 15.0 m at 53° below the positive x-axis.

### Step 2: Determine the Components of Vectors
First, let's break down vectors \(\mathbf{A}\) and \(\mathbf{B}\) into their x and y components:

#### Components of \(\mathbf{A}\)
- \(A_x = A \cos(180° - 30°) = 8.0 \cos(150°)\)
- \(A_y = A \sin(180° - 30°) = 8.0 \sin(150°)\)

Substituting:
\[
A_x = 8.0 \times \cos(150°) = 8.0 \times (-\sqrt{3}/2) \approx -6.93 \, \text{m}
\]
\[
A_y = 8.0 \times \sin(150°) = 8.0 \times (1/2) = 4.0 \, \text{m}
\]

#### Components of \(\mathbf{B}\)
- \(B_x = B \cos(53°) = 15.0 \cos(53°)\)
- \(B_y = B \sin(-53°) = 15.0 \sin(-53°)\)

Substituting:
\[
B_x = 15.0 \times \cos(53°) \approx 15.0 \times 0.6 = 9.0 \, \text{m}
\]
\[
B_y = 15.0 \times \sin(-53°) \approx 15.0 \times (-0.8) = -12.0 \, \text{m}
\]

### Step 3: Calculate the Components of \(-\mathbf{A} - \mathbf{B}\)
Now compute the components of the vector \(-\mathbf{A} - \mathbf{B}\):

- \(-A_x = -(-6.93) = 6.93\)
- \(-A_y = -(4.0) = -4.0\)

Thus:
\[
(-A - B)_x = (-A_x) - B_x = 6.93 - 9.0 = -2.07 \, \text{m}
\]
\[
(-A - B)_y = (-A_y) - B_y = -4.0 - (-12.0) = 8.0 \, \text{m}
\]

### Step 4: Find the Magnitude of \(-\mathbf{A} - \mathbf{B}\)
The magnitude is calculated using the Pythagorean theorem:

\[
|-\mathbf{A} - \mathbf{B}| = \sqrt{((-A - B)_x)^2 + ((-A - B)_y)^2}
\]

Substitute the values:

\[
|-\mathbf{A} - \mathbf{B}| = \sqrt{(-2.07)^2 + (8.0)^2} \approx \sqrt{4.28 + 64} = \sqrt{68.28} \approx 8.27 \, \text{m}
\]

### Step 5: Find the Direction of \(-\mathbf{A} - \mathbf{B}\)
The direction is given by:

\[
\theta = \tan^{-1}\left(\frac{(-A - B)_y}{(-A - B)_x}\right)
\]

Substitute the values:

\[
\theta = \tan^{-1}\left(\frac{8.0}{-2.07}\right) \approx \tan^{-1}(-3.86) \approx -75° 
\]

Since the angle is measured counterclockwise from the positive x-axis, and it is in the second quadrant, the direction is \(180 - 75 = 105°\).

### Final Answers
1. **Magnitude of \(-\mathbf{A} - \mathbf{B}\):** \(8.27 \, \text{m}\)
2. **Direction of \(-\mathbf{A} - \mathbf{B}\):** \(105°\) (measured counterclockwise from the positive x-axis)

Would you like further details on any of these steps or have any other questions?

### Follow-up Questions
1. How do we calculate vector components in different quadrants?
2. What is the significance of negative angles in vector calculations?
3. How to find the unit vector in the direction of \(-\mathbf{A} - \mathbf{B}\)?
4. What is the process for finding the resultant vector of multiple vectors?
5. How would the calculations change if vectors \(\mathbf{C}\) and \(\mathbf{D}\) were involved?

#### Tip
Always sketch vectors and their components to visualize their directions and magnitudes before performing calculations.

Use your answers to find the magnitude of -A - B and the direction of -A - B given vectors A and B in the figure.

Solve the problem of determining the magnitude and direction of -A - B using vector components and trigonometric formulas. This guide covers key concepts in vector subtraction, trigonometry, and the Pythagorean theorem.

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