Here is an apparently simple geometric problem :
Here are the hypothesis : $\theta$, $OC$, $OS$ and $JI$ are known lengths/angles. Furthermore the lines $(OS)$ and the blue one going by $I$ are parallel and $(JI)$ is orthogonal with $(OS)$. The angles $\widehat {JIS'}$ and $\widehat {SIJ}$ have the same value $\alpha$.
$\alpha$ is not known.
The goal is to find the distance of the orange path (in plain line), namely $SI+IC$.
What I have done :
- it is easy to see that the triangle $SIS'$ is isoscele,
- I have managed to compute the distance $CS$, which is :$\left( (OS-OC \cos \theta)^2+OC^2\sin^2 \theta \right)^{1/2}$, so it can be considered known as well.
I made a lot of constructions to try to apply Thales' theorem or something like this but with no success.
Thank you for any help/hint.
As $\alpha$ is not known, it would be worth calculating this in terms of the known lines and angles.
Looking at the figure vertically, we have $$\begin{align}SI\cos\alpha=\frac{l}{2}\\IC\cos\alpha+OC\sin\theta=\frac{l}{2}\\\Rightarrow\cos\alpha=\frac{l-OC\sin\theta}{SI+IC}\end{align}$$ Now examining the figure horizontally, we have $$OC\cos\theta+IC\sin\alpha+SI\sin\alpha=OS\\\Rightarrow\sin\alpha=\frac{OS-OC\cos\theta}{SI+IC}$$ Dividing the above two equations leads to $$\alpha=\arctan\left(\frac{OS-OC\cos\theta}{l-OC\sin\theta}\right)$$ Now that we know $\alpha$, it is straightforward to evaluate the desired length:- $$SI+IC=\frac{l-OC\sin\theta}{\cos\left(\arctan\left(\frac{OS-OC\cos\theta}{l-OC\sin\theta}\right)\right)}$$