Arc Radius Calculation from 2 points 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!
 A: 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 $t=2.24987$.
A: Today I have been attempting a similar thing, as part of a Texas Calculator program I'm writing. I spent several hours trying to figure out how best to calculate this.
So I know the chord length ($C$) and the arc length ($A$). In my case I want to work in degrees, not radians, so I reduced the formula for the radius ($R$) to:
$$
R=\frac{C}{2\sin\left(\frac{360A}{4\pi R}\right)}
$$
But there's the problem of having $R$ on both sides, with no straightforward way to simplify. 
I discovered that my TI-84 Plus calculator has an iterative solver built-in, so I can modify the formula slightly so that it equals zero, and plug it in:
$$
0=\frac{C}{2\sin\left(\frac{360A}{4\pi R}\right)}-R
$$
Or in TI Basic it's:
C/(2sin(360A/(4πR)))-R

I can set my $A$ and $C$ variables, and then put that into the solve function, like this:
19→A
17→C
solve(C/(2sin(360A/(4Rπ)))-R,R,.5C,{.5C,10^10})→R

The crucial part of this is that you only want to search for solutions starting from $\frac{1}{2}C$, as in the real world you can't have a radius smaller than this. I set $10^{10}$ as the maximum.
I believe you can do similar iterative solving within Excel, or within a programming language of your choice. The amount of precision you can get depends on the implementation.
