How to solve an Hyperbolic triangle when all is given except angle C and side c) Another Hyperbolic triangle problem (all given except angle  $\angle C$, and side $c$)
I thought that after asking How to solve an hyperbolic Angle Side Angle triangle? I could solve all hyperbolic triangles, but I am still stumped with SSA and AAS from both you can via the hyperbolic law of sinus calculate one extra value to get to AASS but then I got stuck again therefore: 
If from an hyperbolic triangle $ \triangle ABC$ the angles  $\angle A$, $\angle B$ and sides $a$ and $b$ are given. (so a combination of ASS or AAS )
How can I calculate the angle $\angle C$ or side $c$?
I thought with the two variations of the Hyperbolic law of cosines ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )
or the hyperblic law of sinus ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )  
It must be easy, but neiter law can solve this problem.
Or am I overlooking something (obious) again?
 A: Somehow the following formula's work:
$$ \tanh\left(\frac{c}{2}\right) = \tanh\left(\frac{1}{2} (a-b)\right) \frac{\sin{\left(\frac{1}{2}(A+B)\right)}}{\sin{\left(\frac{1}{2}(A-B)\right)}} $$ 
and
$$ \tan\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{1}{2} (A-B)\right)} \frac{\sinh\left(\frac{1}{2}(a-b)\right)}{\sinh\left(\frac{1}{2}(a+b)\right)} $$ 
They are hyperbolic alternatives to the spherical formulas mentioned at https://en.wikipedia.org/wiki/Solution_of_triangles#A_side.2C_one_adjacent_angle_and_the_opposite_angle_given
(maybe they can be simplified) 
I don't know exactly why they work (and i won't accept this answer till i know) 
user MvG advised me to investigate
http://en.wikipedia.org/wiki/Spherical_trigonometry#Napier.27s_analogies
http://en.wikipedia.org/wiki/Tangent_half-angle_formula
and
http://en.wikipedia.org/wiki/Hyperbolic_function#Hyperbolic_functions_for_complex_numbers
so maybe after that I can explain them :) 
proof based on the proof for spherical geometry by I Hothunter, section 52 
(http://www.gutenberg.org/ebooks/19770 )
But then rewritten for hyperbolic geometry 
The hyperbolic sinus rule ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )
$$ \frac{\sin A}{sinh(a)} = \frac{\sin B}{sinh(b)}= \frac{\sin C}{sinh(c)} $$
from it follows that 
$ \frac{\sin A + \sin B}{sinh(a) + sinh(b)} = \frac{\sin C}{sinh(c)} $
and 
$ \frac{\sin A - \sin B}{sinh(a) - sinh(b)} = \frac{\sin C}{sinh(c)} $
In the following $ \frac{\sin C}{sinh(c)} $ or any of the other equivalents is $ SQ $ 
The hyperbolic cosinus rule ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )
has two forms
(CR1) $ cosh(a) = cosh (b) cosh(c) - sinh(b) sinh(c) \cos A $
and
(CR2) $ \cos A =  - \cos B \cos C + \sin B \sin C cosh(a) $
CR2 rewritten
(1) $  \sin B \sin C cosh(a) = \cos A  + \cos B \cos C $
and for $ \angle B $
(2) $  \sin A \sin C cosh(b) = \cos B  + \cos A \cos C $
1  and 2 involving SQ
(3) $  SQ \sin C cosh(a) sinh(b) = \cos A  + \cos B \cos C $
(4) $  SQ \sin C cosh(b) sinh(a) = \cos B  + \cos A \cos C $
adding 3 and 4 together
(5) $  SQ \sin C ( cosh(a) sinh(b) + cosh(b) sinh(a) ) = (\cos B  + \cos A ) * (1 + \cos C )$
(6) $ sinh(x + y) = sinh (x) cosh (y) + cosh (x) sinh (y) $
therefore 
(7) $  SQ \sin C sinh(a+b) = (\cos B  + \cos A ) (1 + \cos C )$
(8) $ \frac {\sin X}{1 + \cos X } = \tan (\frac {X}{2}) $
therefore
(9) $  SQ \tan (\frac {C}{2}) sinh(a+b) = (\cos B  + \cos A ) $
(10) $  SQ = (\frac {\sin A  + \sin B }{ sinh(a)  + sinh(b)} $ 
therefore
(11) $  (\sin A  + \sin B ) \tan (\frac {C}{2}) sinh(a+b) = (\cos B  + \cos A ) (sinh(a)  + sinh(b))$
(12) $ \tan(\frac {1}{2} (A + B) ) = \frac {\sin A  + \sin B}{\cos B  + \cos A} $
(13) $ \tan(\frac {1}{2} (A + B) ) \tan (\frac {C}{2}) sinh(a+b) = (sinh(a)  + sinh(b)) $
(14) $ \tan (\frac {C}{2}) = \frac {1}{\tan(\frac {1}{2} (A + B) ) }  \frac {sinh(a)  + sinh(b)}{sinh(a+b)} $ 
That should do :) 
A: Suppose you know angles $\alpha$ and $\beta$ as well as edge lengths $a$ and $b$. As you said, one of these can be computed from the other three using the law of sines. So you need to find $c$ and $\gamma$.
Pick one of the law of cosines, e.g.
$$\cosh a=\cosh b\cosh c - \sinh b\sinh c\cos \alpha$$
Then combine that with the well-known identity $\cosh^2 c-\sinh^2 c=1$ and eliminate one variable, e.g. $\sinh c$ using e.g. a resultant. You end up with a quadratic equation for the remaining variable, in this case
$$(\sinh^2b\cos^2\alpha - \cosh^2b)\cosh^2c +
(2\cosh a\cosh b)\cosh c -
(\sinh^2b\cos^2\alpha + \cosh^2a)=0$$
As you can see, this is a quadratic equation in $\cosh c$. Compute its two solutions, and check which of them is the one you need. Perhaps one value is outside the range of $\cosh$? To find $\gamma$ you can again use the law of sines.
