Calculate sin(x) over sin(y) I'm given the equation $$\frac{\sin(\delta)}{\sin(102.9^\circ-\delta)}= b\cdot\frac{\sin(\alpha)}{a\sin(\beta)}$$ where I know the values of $a$, $b$, $\alpha$ and $\beta$.
How can I solve the equation for $\delta$?
To expand @Piquito comment your problem can be written as:
$$\sin(\delta)=A \sin(a-\delta)$$
One way to solve such problem is to write $\sin(a-\delta)=\sin(\delta) \cos(a)-\cos(\delta) \sin(a)$ in order to have:
$$(1-A\cos(a))\sin(\delta)=-A\sin(a) \cos(\delta)$$
which is easier to solve.
I would just do Mathematica here. The solutions are non-trivial. The code

FullSimplify[Solve[Sin[delta]/Sin[angle-delta]==b Sin[alpha]/(a Sin[beta]),delta]]

yields

$$\left\{\left\{\delta \to -\cos ^{-1}\left(\frac{-a-b \sin (\alpha ) \cos
   (\text{angle}) \csc (\beta )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a \cos
   (\text{angle})+b \sin (\alpha ) \csc (\beta ))}}\right)\right\},\left\{\delta \to
   \cos ^{-1}\left(\frac{-a-b \sin (\alpha ) \cos (\text{angle}) \csc (\beta
   )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a \cos (\text{angle})+b \sin
   (\alpha ) \csc (\beta ))}}\right)\right\},\left\{\delta \to -\cos
   ^{-1}\left(\frac{a+b \sin (\alpha ) \cos (\text{angle}) \csc (\beta )}{\sqrt{a^2+b
   \sin (\alpha ) \csc (\beta ) (2 a \cos (\text{angle})+b \sin (\alpha ) \csc (\beta
   ))}}\right)\right\},\left\{\delta \to \cos ^{-1}\left(\frac{a+b \sin (\alpha )
   \cos (\text{angle}) \csc (\beta )}{\sqrt{a^2+b \sin (\alpha ) \csc (\beta ) (2 a
   \cos (\text{angle})+b \sin (\alpha ) \csc (\beta ))}}\right)\right\}\right\}$$
Here the "angle" is, of course, your $102.9^{\circ}$.