Answer
(a) By a factor of 2.
(b) By a factor of one. That is, the angle remains the same.
Work Step by Step
The vector $A$ has components $A_{x}$ and $A_{y}$. So:
A $=\sqrt {(A_{x})^{2}+(A_{y})^{2}}$ and $\theta=tan^{-1}(\frac{A_y}{A_x})$
If $A_{x}$ and $A_{y}$ are doubled:
(a) A' $=\sqrt {(2A_{x})^{2}+(2A_{y})^{2}}=\sqrt {4(A_{x})^{2}+4(A_{y})^{2}}=2\sqrt {(A_{x})^{2}+(A_{y})^{2}}$.
A' = 2A
(b) $\theta~'=tan^{-1}(\frac{2A_y}{2A_x})=tan^{-1}(\frac{A_y}{A_x})=\theta$
