We have the following triangle in which $BC=1000$, $AC=400$, angle $B$ is $45$ degrees, and angle $A$ is $\theta$.
I have to find the angle $\theta$, so I use sine law as follows:
$$\frac{1000}{sin(\theta)}=\frac{400}{sin(45)}$$
$$\therefore\ sin(\theta)=\frac{1000}{400\sqrt{2}}$$
$$\therefore\ sin(\theta)=\frac{5}{2\sqrt{2}}=1.767766953$$
Which is not possible! But where is the mistake?
GReyes is right that your assumed side lengths do not make a triangle -- but that in itself does not explain your troubles, because your calculation doesn't even use the length of $AB$ anyway. (Note: this value for $AB$ was later edited out of the question.)
The real problem is that when $AC<BC$, then the angle at $B$ cannot be arbitrarily large. You need it to be small enough that a line from $B$ that makes the desired angle with $BC$ actually intersects a circle around $C$ with the specified radius $AC$. It's easy to see geometrically that $90^\circ$ will always be too large -- and here where $AC:BC = 4:10$, even $45^\circ$ is too large.