Derive this trigonometric result? I'm revising for an exam tomorrow, but got stuck on the omission of a derivation from this:

This is an image of a triangulation sensor, I worked out how they derived r, but I've no idea where to begin with the derivation for beta. Could anyone shed some light on this for me?
Thanks
 A: Hint: Use sine theorem. I'll post the complete answer when I get home.
I seem to be getting some problems with that $-1$ constant after $\arctan$. Are you sure those are the right formulas?
The idea to get angle $\beta$ in terms of $\alpha_1,\alpha_2$ is the following:
Apply sine theorem in the two halves of triangle and obtain 
$\displaystyle \frac{\sin(\beta+\alpha_1+\alpha_2)}{\sin \beta}=\frac{\sin \alpha_2}{\sin\alpha_1}$. From here, using usual trigonometric formulas, you can extract $\tan \beta$, and then try and prove it has a closed form as presented in your formulas.
Here are some more details. We have
$\frac{\sin \beta \cos(\alpha_1+\alpha_2)+\cos\beta\sin(\alpha_1+\alpha_2)}{\sin\beta}=\frac{\sin\alpha_2}{\sin\alpha_1}$.
Then $\cos(\alpha_1+\alpha_2)+\cot\beta\sin(\alpha_1+\alpha_2)=\frac{\sin\alpha_2}{\sin\alpha_1}$. From here, you get a formula for $\cot \beta$, then you invert and get $\tan \beta$. I didn't manage to get your closed form, but if there is a way, this is the path. :)
