To find the $x$- and $y$-components of the vectors $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, and $\mathbf{D}$, we will use trigonometry. Each vector's components can be found using the following formulas:

• For a vector with magnitude $r$ and angle $\theta$ from the x-axis: $x \text{-component} = r \cos(\theta)$ $y \text{-component} = r \sin(\theta)$

Given the directions and angles, let's break down each vector:

Vector $\mathbf{A}$:

• Length: 8.00 meters
• Direction: Along the negative $y$-axis
• Angle: $270^\circ$ (measured from the positive $x$-axis)

$\mathbf{A}_x = 8.00 \cos(270^\circ) = 8.00 \times 0 = 0.00 \text{ m}$ $\mathbf{A}_y = 8.00 \sin(270^\circ) = 8.00 \times (-1) = -8.00 \text{ m}$

Vector $\mathbf{B}$:

• Length: 15.0 meters
• Direction: Makes a $30.0^\circ$ angle with the positive $y$-axis
• Angle: $90^\circ - 30^\circ = 60^\circ$ (relative to the positive $x$-axis)

$\mathbf{B}_x = 15.0 \cos(60^\circ) = 15.0 \times 0.5 = 7.50 \text{ m}$ $\mathbf{B}_y = 15.0 \sin(60^\circ) = 15.0 \times \frac{\sqrt{3}}{2} \approx 12.99 \text{ m}$

Vector $\mathbf{C}$:

• Length: 12.0 meters
• Direction: Makes a $25.0^\circ$ angle with the negative $x$-axis
• Angle: $180^\circ + 25^\circ = 205^\circ$ (relative to the positive $x$-axis)

$\mathbf{C}_x = 12.0 \cos(205^\circ) = 12.0 \times \cos(205^\circ) \approx -10.88 \text{ m}$ [ \mathbf{C}_y = 12.0 \sin(205^\circ) = 12.0 \times \sin(205^\circ) \approx -5.07 \text