hyperbolic trigonometry when one angle =0 The hyperbolic trigonometry functions don't really help when you have  one angle =0 (the remaining lenght of side $AB$ becomes ${\infty}-{\infty}$ )
Given a triangle $\triangle AB \Omega$ with $\angle \Omega =0 $ then by this fact alone we get $ A\Omega = \infty , B\Omega = \infty $
When one of the angles $\angle A , \angle B $ is right the remaining side and angle are related by the formula's for the angle of parallelism.
But then what if we have a $\triangle AB \Omega$ with $\angle \Omega =0 ,A\Omega = \infty , B\Omega = \infty $  and neither  $\angle A , \angle B $ is right, How can we calculate the length of side $AB$? 
 A: Look at the hyperbolic law of cosines. It tells you that
$$\cosh c=\frac{\cos\alpha\cos\beta+\cos\gamma}{\sin\alpha\sin\beta}$$
Now in your case, $\gamma=0$ so you have
$$\cosh c = \frac{\cos\alpha\cos\beta+1}{\sin\alpha\sin\beta}$$
In the case of $\beta=\frac{\pi}{2}$ you get $\cosh c=\frac{1}{\sin\alpha}$ or equivalently $\sin\alpha=\frac{1}{\cosh c}$ which is one of the formulas for the angle of parallelism, just as you said.

For some reason, I had read the law of cosine formula incorrectly when I first tried to answer, and therefore took an immense detour. To answer your question, it isn't neccessary, but I'll leave it in place since it's a technique which might be useful for other applications.
One thing you can do is work with a model. I'd use the Poincaré half plane for this. Without loss of generality, $\Omega$ is the point at infinity, so the geodesics of infinite length would be Euclidean vertical lines. $A=(0,1)$ still without loss of generality. So now you have a point $B=(x,y)$ which determines both the angles and the distance between $A$ and $B$.
What's the equation of the line $AB$? It's a circle with center on the $x$ axis, e.g. at $C=(c,0)$, such that
\begin{align*}
c^2+1^2&=(c-x)^2+y^2\\
c^2+1&=c^2-2cx+x^2+y^2\\
2cx&=x^2+y^2-1\\
c&=\frac{x^2+y^2-1}{2x}
\end{align*}
Then you have $\overrightarrow{CA}$ and $\overrightarrow{CB}$ as radial directions, and orthogonal to these the tangential directions. The angle the tangential directions make with the vertical axis are your corner angles of the triangle. They are equal to the angle the radial directions make with the horizontal axis. So up to some sign mistake I might have made, you have
$$\tan\alpha=\frac{1}{-c}=\frac{2x}{1-x^2-y^2}\qquad
\tan\beta=\frac{y}{x-c}=\frac{2xy}{1+x^2-y^2}$$
If you know both angles, you can use these equations to find $x$ and $y$, and then you can use the distance formula
$$\cosh d=1+\frac{x^2+(y-1)^2}{2y}$$
Note that the two equations from which you read $x$ and $y$ would both be quadratic, so you'd essentially be intersecting two conics there. You may be able to rule out some solutions, e.g. with $y<0$. But since two conics might theoretically intersect in up to four points, this approach does not easily give a proof that two angles uniquely define the remaining length.
So it's good that I had another look at the law of cosines, and came up wth the far simpler approach I mentioned up front.
