Let $\overrightarrow{C}=\left(3.15\text{m},{15}^{°}\right.$above the negative $x$-axis ) and $\overrightarrow{D}=$$\left(25.6\text{m},{30}^{°}\right.$ to the right of the negative $y$-axis). Find the $x$-and -$y$components of each vector.

Vector can be defined as a physical quantity which has magnitude as well as direction.

Draw the diagram for the given vector $\overrightarrow{C}$as follows,

Here, the vector $\overrightarrow{C}$is above the negative $x$-axis as shown in the figure.

Calculate the magnitude of the $x$-component, and $y$-component of the vector $\overrightarrow{C}$.

The magnitude of$\overrightarrow{C}$is,

$C=3.15\text{m}$

The$x$-component of$\overrightarrow{C}$is,

$C_x=-C\cos \theta$

Here, $C$is the magnitude of the vector $\text{C}$and $\theta$is the angle.

Substitute $3.15\text{m}$ for $C$and ${15}^{°}$for $\theta$in above equation.

$\begin{array}{l}{C}_x=-(3.15\text{m})\cos \left({15}^{°}\right)\\ =-(3.15\text{m})\left(0.9659\right)\\ =-3.04\text{m}\end{array}$

Therefore, the magnitude of the $x$-component of the vector $\overrightarrow{C}$is$-3.04\text{m}$.

The $y$-component of the vector $\overrightarrow{C}$is,

$C_y=C\sin \theta$

Substitute $3.15\text{m}$for $C$and ${15}^{°}$for $\theta$in above equation.

$\begin{array}{l}{C}_y=(3.15\text{m})\sin \left({15}^0\right)\\ =(3.15\text{m})\left(0.2588\right)\\ =0.815\text{m}\end{array}$

Therefore, the magnitude of the $y$-component of the vector $\overrightarrow{C}$is$0.815\text{m}$.

Draw the vector diagram for the given vector $\overrightarrow{D}$as follows,

Calculate the magnitudes of the$x$-component, and$y$-component of the vector$\overrightarrow{D}$.

The magnitude of$\overrightarrow{D}$is,

$D=25.6\text{m}$

The$x$-component of$\overrightarrow{D}$is,

$D_x=D\sin \theta$

Here,$D$is the magnitude of the vector$\overrightarrow{D}$and$\theta$is the angle made by the vector$\overrightarrow{D}$on the right side of the negative $y$-axis.

Substitute$25.6\text{m}$for$D$and${30}^{°}$for$\theta$in above equation.

$\begin{array}{l}{D}_x=(25.6\text{m})\sin \left({30}^{°}\right)\\ =(25.6\text{m})\left(0.5\right)\\ =12.8\text{m}\end{array}$

Therefore, the magnitude of the $x$-component of the vector $\overrightarrow{D}$is$12.8\text{m}$.

The$y$-component of the vector$\overrightarrow{D}$is,

$D_y=-D\cos \theta$

Substitute $25.6\text{m}$for $D$and ${30}^{°}$for $\theta$in above equation.

$\begin{array}{l}{D}_y=-(25.6\text{m})\cos \left({30}^{°}\right)\\ =-(25.6\text{m})\left(0.86602\right)\\ =-22.17\text{m}\end{array}$

Therefore, the magnitude of the $y$-component of the vector $\overrightarrow{D}$is $-22.17\text{m}$.