How to solve this spherical trigonometry situation with missing information?

Suppose $b$ and $c$ were given constants. If $C = \theta - y$ (where $\theta$ is given) and $a = 90 \deg - y $, is it possible to solve for $y$? It seems like there must be a solution, since $y$ can't just be any arbitrary value or else the triangle won't be valid.

So far I have used cosine law for spherical surfaces:

$sin(a) = cos(b)cos(c) + sin(b)sin(c)cos(A)$

$cos(y) = cos(b)cos(c) + sin(b)sin(c)cos(A)$

and sine law:

$ \frac{sin(B)}{cos(b)} = \frac{sin(\theta - y)}{sin(c)} = \frac{sin(A)}{sin(a)} $

$ \frac{sin(B)}{cos(b)} = \frac{sin(\theta - y)}{sin(c)} = \frac{sin(A)}{cos(y)} $

But I can't get it to an expression of $y$. I also cannot think of a way to solve this even if this was a non-spherical trigonometry problem.

There's not a nice solution to this system. But briefly, we have $$ \sin y = \cos b \cos c + \sin b \sin c \cos A $$ (note that you didn't quite apply the law of cosines right) and $$ \sin A = \frac{\sin(\theta - y) \cos y}{\sin c}. $$ But since $\cos A = \pm \sqrt{1 - \sin^2 A}$, we can combine these equations to yield $$ \sin y = \cos b \cos c + \sin b \sqrt{ \sin^2 c - \sin^2(\theta - y) \cos^2 y}. $$ This equation implicitly defines $y$. For any given values of $b$ and $c$, you could plot both sides and see where they intersect to find $y$.

In principle, you could also rearrange this last equation, square both sides a couple of times, and repeatedly use the identities $\cos^2 y = 1 - \sin^2 y$ to reduce this to a polynomial in $\cos y$ or $\sin y$. However, it looks to me that this polynomial will be of fairly high order, meaning that it's unlikely to have a nice closed-form solution if it has one at all.