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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 $\delta$?

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    $\begingroup$ Do you have the equation $\sin(x)=A\sin(a-x)$ where $a$ and $A$ are constants. $\endgroup$ – Piquito Apr 16 '18 at 17:15
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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.

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    $\begingroup$ To solve the problem in the last form you can for example write $\tan(\delta)=\frac{A \sin(a)}{A\cos(a)-1}$ and take the $\arctan$. $\endgroup$ – Delta-u Apr 16 '18 at 17:21
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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 $102.9^{\circ}$.

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