I've seen other questions on here and tried to follow them, but I was hoping somebody could help me understand where I'm going wrong in my solution, and point me in the right direction.
I've got the points shown in the above picture: $A$ and $B$, which are known - thus the distance between them (labeled $\overline{\rm AB}$). Also known/given is the arc length, $\overset{\mmlToken{mo}{⏜}}{AB\,}$. I am trying to find R ($\overline{\rm AC}$ or $\overline{\rm BC}$) and/or $\theta$. (C is unknown.)
$$\overset{\mmlToken{mo}{⏜}}{AB\,} = \theta R,\space \text{thus}\space R = \frac{\overset{\mmlToken{mo}{⏜}}{AB\,}}{\theta} $$
$$2R \sin (\frac{\theta}{2}) = \overline{\rm AB}, \space \text{so}\space R = \frac{\overline{\rm AB}}{2\sin (\frac{\theta}{2})}$$
Using these, we can find:
$$\frac{\overset{\mmlToken{mo}{⏜}}{AB\,}}{\theta} = \frac{\overline{\rm AB}}{2\sin (\frac{\theta}{2})}$$
With some rearranging,
$$\sin (\frac{\theta}{2}) = \frac{\overline{\rm AB}}{\overset{\mmlToken{mo}{⏜}}{AB\,}}\frac{\theta}{2}$$
From here, we can do some variable reassignment: I'll say $t = \frac{\theta}{2}$, and $k = \frac{\overline{\rm AB}}{\overset{\mmlToken{mo}{⏜}}{AB\,}}$.
This gives us $\sin (t) = kt$
I see no way for $k$ to be $>1$ (and $k=1$ only if the radius is infinite and theta is $0$), but I suppose theta could be basically any angle - I'm solving for positive, and for my use case I expect it will always fall in the $0-\pi$ range, but I guess it doesn't have to.
I feel fairly confident on the geometry side of things; that makes sense to me. I haven't done much by way of approximation, so I don't know where to go from here. I understand from my reading of other questions that this is a "transcendental" equation, which apparently means something like "doesn't have a closed-form algebraic solution"? Thus my question: have I screwed anything up? If so, what? If not, where do I go from here?
I tried doing something like $\frac{\sin (t)}{t} = k$, but I still don't know how to computationally approximate this - I'm embedding this in an algorithm I'm using on a website for solving a specific class of geometry problems, so I'd like it to be as accurate as possible - an arbitrary number of decimal places would be great, but I'll settle for like $4$.
Thank you!
In order to solve for $t$
$$\sin(t)=k t$$ you could use the $\large 1400$ years old approximation $$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad\text{for} \qquad0\leq t\leq\pi$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
Using it, you just need to solve the quadratic equation itn $t$ $$4k t^2-4 (\pi k-4) t-\pi (16-5 \pi k)=0$$ which gives $$t=\frac{2 \sqrt{-\pi ^2 k^2+2 \pi k+4}+\pi k-4}{2 k}$$
Using $k=0.3456$, this would give $t=2.24911$ while the solution given by Newton method is $2.25049$ which is not too bad. Now, you can polish the root using Newton method which will generate the following iterates (the ridiculous nmber figures being on purpose) $$\left( \begin{array}{cc} n & t_n \\ 0 & 2.2491058627233529704 \\ 1 & 2.2504866087065066172 \\ 2 & 2.2504858470778471286 \\ 3 & 2.2504858470776155599 \end{array} \right)$$
Edit
If the angle is not very close to $0$ or $\pi$, a good approximation is $$\frac{\sin(t)}t=\frac{\frac{2}{\pi }-\frac{2}{3 \left(\pi ^2-8\right)}\left(t-\frac{\pi }{2}\right)+\frac{48-5 \pi ^2}{6 \pi \left(\pi ^2-8\right)}\left(t-\frac{\pi }{2}\right)^2 } {1+\frac{5 \pi ^2-48}{3 \pi \left(\pi ^2-8\right)}\left(t-\frac{\pi }{2}\right)+\frac{1}{12}\left(t-\frac{\pi }{2}\right)^2 }$$ which is still a quadratic equation in $\left(t-\frac{\pi }{2}\right)$.
For the worked case, it gives $t=2.24987$.