Answer
If $a$ is twice as long as $b$, $A$ is not necessarily twice as large as $B$.
Work Step by Step
The law of sines for a triangle having sides $a, b$ and angles $A, B$:
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
This would mean
$$\frac{a}{b}=\frac{\sin A}{\sin B}$$
Since $a$ is twice as long as $b$:
$$\frac{a}{b}=\frac{\sin A}{\sin B}=2$$
Therefore, $$\sin A=2\sin B$$
Now this only means the value of $\sin A$ is twice as large as that of $\sin B$, not necessarily angle $A$ is twice as large as angle $B$.
To show this, we can take a case where $\sin B=\frac{1}{2}$ and $\sin A=1$.
Obviously, $\sin A=2\sin B$ because $1=2\times\frac{1}{2}$
However, $\sin A=1$ refers to $A=90^\circ$, while $\sin B=\frac{1}{2}$ refers to $B=30^\circ$. Yet here we can see that $A=3B$.
Therefore, if $a$ is twice as long as $b$, only the value of $\sin A$ is necessarily twice as large as that of $\sin B$. $A$ is not necessarily twice as large as $B$.
