[Math] Finding the intersection of two lines, in polar coordinates

trigonometry

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.

Best Answer

HINT

We have

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

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

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

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

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