Answer
$$ \vec C =-3.04 \;{\rm m}\;\hat{i}+0.815 \;{\rm m}\;\hat{j} $$
$$ \vec D =12.8 \;{\rm m} \;\hat{i}-22.2 \;{\rm m}\;\hat{j} $$
Work Step by Step
We know that the horizontal component of any vector is given by
$$C_x=|C|\cos\theta$$ whereas $\theta$ is the angle between the vector and the positive $x$-axis.
The angle that vector $\vec C$ makes with $+x$-axis is given by
$$\theta_{\vec C}=180^\circ -15^\circ=\color{blue}{\bf 165^\circ}$$
Thus,
$$C_x=|C|\cos\theta_{\vec C}=3.15\cos 165^\circ=\color{red}{\bf-3.04}\;\rm m$$
And its vertical component is given by
$$C_y=|C|\sin\theta_{\vec C}=3.15\sin 165^\circ=\color{red}{\bf 0.815}\;\rm m$$
$$\boxed{ \vec C =-3.04 \;{\rm m} \;\hat{i}+0.815 \;{\rm m}\;\hat{j} }$$
Now we need to find the angle that vector $\vec D$ makes with $+x$-axis:
$$\theta_{\vec D}=270^\circ +30^\circ=\color{blue}{\bf 300^\circ}$$
Thus, its horizontal component is given by
$$D_x=|D|\cos\theta_{\vec D}=25.6\cos 300^\circ=\color{red}{\bf12.8}\;\rm m$$
And its vertical component is given by
$$D_x=|D|\sin \theta_{\vec D}=25.6\sin 300^\circ=\color{red}{\bf-22.2}\;\rm m$$
$$\boxed{ \vec D =12.8 \;{\rm m} \;\hat{i}-22.2 \;{\rm m}\;\hat{j} }$$
