II Let \( \vec{C}=\left(3.15 \mathrm{~m}, 15^{\circ}\right. \) above the negative \( x \)-axis \( ) \) and \( \vec{D}= \) \( \left(25.6 \mathrm{~m}, 30^{\circ}\right. \) to the right of the negative \( y \)-axis \( ) \). Find the \( x \) - and \( y \)-components of each vector.

To find the components of the vectors \( \vec{C} \) and \( \vec{D} \), we first need to break them down into their respective \( x \)- and \( y \)-components using trigonometric functions. For \( \vec{C} \) at an angle of \( 15^\circ \) above the negative \( x \)-axis: - The angle with the positive \( x \)-axis is \( 180^\circ - 15^\circ = 165^\circ \). - The \( x \)-component is: \[ C_x = 3.15 \cos(165^\circ) \approx 3.15 \cdot (-0.9659) \approx -3.04 \, \text{m} \] - The \( y \)-component is: \[ C_y = 3.15 \sin(165^\circ) \approx 3.15 \cdot 0.2588 \approx 0.82 \, \text{m} \] For \( \vec{D} \) at an angle of \( 30^\circ \) to the right of the negative \( y \)-axis: - The angle with the positive \( y \)-axis is \( 90^\circ - 30^\circ = 60^\circ \). - The angle with the positive \( x \)-axis is \( 180^\circ - 60^\circ = 120^\circ \). - The \( x \)-component is: \[ D_x = 25.6 \cos(120^\circ) \approx 25.6 \cdot (-0.5) \approx -12.8 \, \text{m} \] - The \( y \)-component is: \[ D_y = 25.6 \sin(120^\circ) \approx 25.6 \cdot \left(\frac{\sqrt{3}}{2}\right) \approx 25.6 \cdot 0.8660 \approx 22.19 \, \text{m} \] In summary: - For \( \vec{C} \): \( C_x \approx -3.04 \, \text{m}, C_y \approx 0.82 \, \text{m} \) - For \( \vec{D} \): \( D_x \approx -12.8 \, \text{m}, D_y \approx 22.19 \, \text{m} \)