Finding the intersection of two lines, in polar coordinates The sticking point is figuring out the substitutions for a ratio of cosines of differences.
I have a pair of lines in polar coords:

$$r = \frac{s_1}{\cos(\theta - \alpha_1)} \qquad r = \frac{s_2}{\cos(\theta - \alpha_2)}$$

where

$$\begin{align}
\alpha_1 &= 6^\circ \\
s_1 &= 0.9945218953682733 \\
\alpha_2 &= 74^\circ \\
s_2 &= 0.27563735581699916
\end{align}$$

I then need to do trigonometric substitution to solve for $\theta$.

$$\begin{align}
\frac{s_1}{\cos(\theta - \alpha_1)} &= \frac{s_2}{\cos(\theta - \alpha_2)} \\[4pt]
\frac{s_1}{s_2} &= \frac{\cos(\theta - \alpha_1)}{\cos(\theta - \alpha_2)} 
\end{align}$$

I am stumped after that point.

Following @gimusi: hint
$$s^1_{val} \cos \theta\cos (s^2_{ang})+s^1_{val} \sin \theta\sin (s^2_{ang})=s^2_{val} \sin \theta\cos (s^1_{ang})+s^2_{val} \sin \theta\sin (s^1_{ang})$$
$$s^1_{val} \cos \theta\cos (s^2_{ang})-s^2_{val} \cos \theta\cos (s^1_{ang})=s^2_{val} \sin \theta\sin (s^1_{ang})-s^1_{val} \sin \theta\sin (s^2_{ang})$$
$$ \cos \theta[s^1_{val}\cos (s^2_{ang})-s^2_{val}\cos (s^1_{ang})]= \sin \theta[s^2_{val} \sin (s^1_{ang})-s^1_{val} \sin (s^2_{ang})]$$
$$ \sin \theta/ \cos \theta=[s^1_{val}\cos (s^2_{ang})-s^2_{val}\cos (s^1_{ang})]/[s^2_{val} \sin (s^1_{ang})-s^1_{val} \sin (s^2_{ang})]$$
$$ \tan \theta=[s^1_{val}\cos (s^2_{ang})-s^2_{val}\cos (s^1_{ang})]/[s^2_{val} \sin (s^1_{ang})-s^1_{val} \sin (s^2_{ang})]$$
$$ \theta=\arctan[[s^1_{val}\cos (s^2_{ang})-s^2_{val}\cos (s^1_{ang})]/[s^2_{val} \sin (s^1_{ang})-s^1_{val} \sin (s^2_{ang})]]$$
then plug into $$r = \frac{s_1}{\cos(\theta - \alpha_1)}$$
which I verified with numpy for my problem!
The numerical answer in this case is $$(\theta=0, r=1.)$$
