Four vectors start at the origin of the x y axes labeled O. Vector A is directed along the negative y axis and has a length of 8.00 meters. Vector B is located in the first quadrant, it makes an angle of 30.0 degrees with the positive y axis and has a length of 15.0 meters. Vector C is located in the third quadrant, it makes an angle of 25.0 degrees with the negative x axis and has a length of 12.0 meters. Vector D is located in the second quadrant, it makes an angle of 53.0 degrees with the negative x axis and has a length of 10.0 meters. - Four vectors starting at the origin

Compute the x- and y-components of the vectors A⃗, B⃗, C⃗, and D⃗ in (Figure 1).

Express your answer in meters.

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

Learn how to compute the x- and y-components of vectors A, B, C, and D starting from their lengths and directions. This problem covers trigonometric concepts and vector decomposition. Suitable for high school students.

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