In most geometry courses, we learn that there's no such thing as "SSA Congruence". In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles $ABC$ and $DEF$ such that $AB = DE$, $BC = EF$, and $\angle A = \angle D$, then we cannot deduce that $ABC$ and $DEF$ are congruent.

However, there are a few special cases in which SSA "works". That is, suppose $ABC$ is a triangle. Let $AB = x$, $BC = y$, and $\angle A = \theta$. For some values of $x$, $y$, and $\theta$, we can uniquely determine the third side, $AC$.

(a) Use the Law of Cosines to derive a quadratic equation in $AC$.

(b) Use the quadratic polynomial you found in part (a) in order to find conditions on $x, y,$ and $\theta$ which guarantee that the side $AC$ is uniquely determined.

I have solved part (a) and gotten $z^2-2xz\cos (\theta) + (x^2-y^2)=0$.

I don't quite get part (b). It is obvious $90^{\circ}$ is a solution but do I have to prove that this is right? Or do I need to prove that this is the only solution? Or is there other answers?

My work: All I know is that the discriminant would be $4x^2z^2\cos^2 \theta-4(-x^2+y^2)$

P.S. My answer for part(a) is correct, right?
In the generic case, there are two triangles (up to congruence).  To see this, suppose we are given AB, BC, and the angle at A.  Fix the segment AB and let BC rotate through a circle (centered at B).  Form the line from A at the given angle from AB.  That line can intersect the circle at two points.

True, it can be tangent to that circle but that's precisely the case in which the leg BC is perpendicular to AC.
You have a quadratic equation for $z$. Most quadratic equations have either two solutions. If you want to uniquely determine z, you need to have only ONE solution, right?

My advice (I'm not going to solve the problem, but this is a pretty good hint!) is to use your knowledge of solving quadratic equations to solve for z, and then ask yourself when the two solutions you get will be equal. For example, if I'm solving $z^2-5z+6$, I could use the quadratic formula to find $z=2$ or $z=3$, so there's two solutions, and I haven't uniquely determined $z$.

If you get stuck, look up "discriminant".