I have a spherical triangle with corners $A,B,C$, angles $\alpha, \beta, \gamma$ and sides $a,b,c$ (which are opposite to the corresponding corners/angles).
I am given $a,b$ (with $a>b$) and $\alpha$. I want to explicitly calculate the other angles and lengths of the triangle by using only the law of sines and the law of cosines for sides and angles.
I am also given a hint, that I should draw a line through point $C$ which intersects side $c$ at right angle (let's call the intersection point $D$). So I want to look at two triangles $A,C,D$ and $B,C,D$ with sides $c', a', b$ and $a, b'', c''$ where $c'+c'' = c$ and $b''=a'$. Their angles are $\alpha, \lrcorner, \gamma'$ and $\beta, \gamma'', \llcorner$ with $\gamma'+\gamma''=\gamma$ and one right angle in each triangle.
My problem is now that especially the law of cosines for sides makes the calculation very complicated and I'm wondering if I'm doing things right.
Starting with the first triangle and the law of sines, I get
$$\sin{a'} = \sin{b}\sin{\alpha}\\
\sin{c'} = \sin{b}\sin{\gamma'}$$
Using the law of cosines for angles, I get $3$ reasonable simple expressions but the law of cosines for sides gives me $3$ terms which look like this
$$\cos{a'}=\cos{c'}\cos{b} + \sin{c'}\sin{b}\sin{\alpha}$$
I tried to solve the resulting system of $8$ equations but it is really complicated and led me to the nonsensical result that a product of sines has to be zero. This is probably due to an algebraic error of mine. And on top of the algebraic problems, I'll also have to take the inverses of trigonometric functions in the end.
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