Finding the intersection of two lines, in polar coordinates

Question

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.

Answer 1

HINT

We have

$$s^1_{val} \cos(\theta - s^2_{ang}) =s^2_{val} \cos(\theta - s^1_{ang})$$

$$s^1_{val} \cos \theta\sin (s^2_{ang})+s^1_{val} \sin \theta\cos (s^2_{ang})=s^2_{val} \cos \theta\sin (s^1_{ang})+s^2_{val} \sin \theta\cos (s^1_{ang})$$

$$s^1_{val} \cos \theta\sin (s^2_{ang})-s^2_{val} \cos \theta\sin (s^1_{ang}) =s^2_{val} \sin \theta\cos (s^1_{ang})-s^1_{val} \sin \theta\cos (s^2_{ang})$$

$$\cos \theta \,[s^1_{val} \sin (s^2_{ang})-s^2_{val} \sin (s^1_{ang})] =\sin \theta [s^2_{val} \cos (s^1_{ang})-s^1_{val} \cos (s^2_{ang})]$$

Answer 2

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 $$(\theta=0, r=1.)$$