For the vector A making an angle θ with the negative X-axis and...

The vector $\vec{A}$ lies in the third quadrant and makes an angle $\theta$ with the negative X-axis.

The component of the vector along the negative X-axis is $A\cos\theta$, so $A_x = -A\cos\theta$.

The component of the vector along the negative Y-axis is $A\sin\theta$, so $A_y = -A\sin\theta$.

Therefore, the vector in standard resolved form is $\vec{A} = -A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$.

Answer: $-A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$