Given the length of an arc and the length of sagitta, can you calculate the angle (radians)?
I struggle to work out all the parameters I need. For instance, to calculate the radius I need the length of the sagitta and the chord length (but I don't have that - to get that I need the radius...)
Do I have too few parameters to get a single answer for this?
On
Satrting from saulspatz's answer, it is sure that, using a good estimate, Newton method would converge in very few iterations provided of a good guess for the root.
The first thing I thought about was to set $\theta=2t$, $k=\frac{s}{a}$ and use the rather good approximation $$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$ This leads to $$t(2 k t^2+2 \pi ^2 k-5 t)=0 \implies \theta_{est}=\frac{5-\sqrt{25-16 \pi ^2 k^2}}{2 k}$$ In order to check it, give $\theta$ a value, compute $k$ and compute $ \theta_{est}$. The table below shows some results $$\left( \begin{array}{ccc} \theta & k & \theta_{est} \\ 0.1 & 0.0124973961 & 0.0987 \\ 0.2 & 0.0249791736 & 0.1974 \\ 0.3 & 0.0374297402 & 0.2962\\ 0.4 & 0.0498335554 & 0.3950 \\ 0.5 & 0.0621751566 & 0.4939 \\ 0.6 & 0.0744391848 & 0.5930 \\ 0.7 & 0.0866104102 & 0.6921 \\ 0.8 & 0.0986737575 & 0.7915 \\ 0.9 & 0.1106143307 & 0.8909 \\ 1.0 & 0.1224174381 & 0.9906 \\ 1.1 & 0.1340686163 & 1.0904 \\ 1.2 & 0.1455536542 & 1.1905 \\ 1.3 & 0.1568586165 & 1.2908 \\ 1.4 & 0.1679698662 & 1.3913 \\ 1.5 & 0.1788740874 & 1.4920 \end{array} \right)$$ and, starting from the estimate $\theta_{0}=\theta_{est}$, Newton iterates $$\theta_{n+1}=\frac{\theta_n \sin \left(\frac{\theta_n }{2}\right)+2 \cos \left(\frac{\theta_n }{2}\right)-2}{\sin \left(\frac{\theta_n }{2}\right)-2 k}$$ would converge in very few iterations.
My next idea was to develop $\frac{1-\cos\left(\frac{\theta}{2}\right)}{\theta}$ as Taylor series built at $\theta=0$ giving $$\frac{1-\cos\left(\frac{\theta}{2}\right)}{\theta}=\frac{\theta }{8}-\frac{\theta ^3}{384}+\frac{\theta ^5}{46080}-\frac{\theta ^7}{10321920}+\frac{\theta ^9}{3715891200}-\frac{\theta ^{11}}{1961990553600}+O\left(\theta ^{13}\right)$$ and to use series reversion to get $$\theta=8 k+\frac{32 k^3}{3}+\frac{1664 k^5}{45}+\frac{159232 k^7}{945}+\frac{4139008 k^9}{4725}+\frac{21929984 k^{11}}{4455}+O\left(k^{13}\right)$$ which could be transformed as a Padé approximant such as $$\theta_{est}=k\,\frac{8 -\frac{13184 }{279}k^2+\frac{200192 }{5859}k^4 } {1-\frac{2020 }{279}k^2+\frac{272512 }{29295}k^4 }$$ Doing the same as before, we should get $$\left( \begin{array}{ccc} \theta & k & \theta_{est} \\ 0.1 & 0.0124973961 & 0.100000000000 \\ 0.2 & 0.0249791736 & 0.200000000000 \\ 0.3 & 0.0374297402 & 0.300000000000 \\ 0.4 & 0.0498335554 & 0.399999999999 \\ 0.5 & 0.0621751566 & 0.499999999991 \\ 0.6 & 0.0744391848 & 0.599999999936 \\ 0.7 & 0.0866104102 & 0.699999999650 \\ 0.8 & 0.0986737575 & 0.799999998468 \\ 0.9 & 0.1106143307 & 0.899999994352 \\ 1.0 & 0.1224174381 & 0.999999981810 \\ 1.1 & 0.1340686163 & 1.099999947499 \\ 1.2 & 0.1455536542 & 1.199999861547 \\ 1.3 & 0.1568586165 & 1.299999661478 \\ 1.4 & 0.1679698662 & 1.399999223847 \\ 1.5 & 0.1788740874 & 1.499998316221 \end{array} \right)$$ which seems to be quite good.
Using the same form, I tried to optimize the coefficients for a better fit over the wole range. What I obtained is $$\theta_{est}=k\,\frac{8-\frac{100815 }{2003}k^2+\frac{26153}{635} k^4}{ 1-\frac{5467 }{717}k^2+\frac{5955}{557}k^4}$$ Reproducing the same table $$\left( \begin{array}{ccc} \theta & k & \theta_{est} \\ 0.1 & 0.0124973961 & 0.0999999999 \\ 0.2 & 0.0249791736 & 0.1999999991 \\ 0.3 & 0.0374297402 & 0.2999999973 \\ 0.4 & 0.0498335554 & 0.3999999947 \\ 0.5 & 0.0621751566 & 0.4999999918 \\ 0.6 & 0.0744391848 & 0.5999999895 \\ 0.7 & 0.0866104102 & 0.6999999885 \\ 0.8 & 0.0986737575 & 0.7999999886 \\ 0.9 & 0.1106143307 & 0.8999999893 \\ 1.0 & 0.1224174381 & 0.9999999891 \\ 1.1 & 0.1340686163 & 1.0999999871 \\ 1.2 & 0.1455536542 & 1.1999999836 \\ 1.3 & 0.1568586165 & 1.2999999814 \\ 1.4 & 0.1679698662 & 1.3999999835 \\ 1.5 & 0.1788740874 & 1.4999999861 \end{array} \right)$$
Update
In order to cover the full range $(0\leq \theta \leq \pi)$ and to get solutions at the price of only a quadratic equation, we can guess the approximation $$\frac{1-\cos\left(\frac{\theta}{2}\right)}{\theta}=\frac{a\theta+b \theta^2 } {1+c\theta+d \theta^2 }$$ and compute the coefficients $a,b,c,d$ in order to have a perfect match at points $\frac \pi 4,\frac \pi 2,\frac {3\pi} 4,\pi$. This gives the nasty values $$a=\frac{52-71 \sqrt{2}+6 \sqrt{10642-7001 \sqrt{2}}}{93 \pi ^2}$$ $$b=\frac{4 \left(26+11 \sqrt{2}-6 \sqrt{754-487 \sqrt{2}}\right)}{93 \pi ^3}$$ $$c=\frac{-325+33 \sqrt{2}+6 \sqrt{2074-167 \sqrt{2}}}{93 \pi }$$ $$d=-\frac{2 \left(-194+30 \sqrt{2}+3 \sqrt{2756-322 \sqrt{2}}\right)}{93 \pi ^2}$$ and the retained solution of the quadratic in $\theta$ is given by $$ \theta_{est}=\frac{\sqrt{(a-c k)^2+4 k (b-d k)}-a+c k}{2 (b-d k)}$$ $$\left( \begin{array}{ccc} \theta & k & \theta_{est} \\ 0.0 & 0.000000 & 0.00000 \\ 0.1 & 0.012497 & 0.09987 \\ 0.2 & 0.024979 & 0.19981 \\ 0.3 & 0.037430 & 0.29980 \\ 0.4 & 0.049834 & 0.39982 \\ 0.5 & 0.062175 & 0.49986 \\ 0.6 & 0.074439 & 0.59991 \\ 0.7 & 0.086610 & 0.69996 \\ 0.8 & 0.098674 & 0.80001 \\ 0.9 & 0.110614 & 0.90005 \\ 1.0 & 0.122417 & 1.00007 \\ 1.1 & 0.134069 & 1.10009 \\ 1.2 & 0.145554 & 1.20009 \\ 1.3 & 0.156859 & 1.30007 \\ 1.4 & 0.167970 & 1.40005 \\ 1.5 & 0.178874 & 1.50002 \\ 1.6 & 0.189558 & 1.59999 \\ 1.7 & 0.200010 & 1.69996 \\ 1.8 & 0.210217 & 1.79993 \\ 1.9 & 0.220167 & 1.89991 \\ 2.0 & 0.229849 & 1.99990 \\ 2.1 & 0.239252 & 2.09990 \\ 2.2 & 0.248365 & 2.19993 \\ 2.3 & 0.257179 & 2.29997 \\ 2.4 & 0.265684 & 2.40003 \\ 2.5 & 0.273871 & 2.50010 \\ 2.6 & 0.281731 & 2.60018 \\ 2.7 & 0.289257 & 2.70026 \\ 2.8 & 0.296440 & 2.80032 \\ 2.9 & 0.303275 & 2.90034 \\ 3.0 & 0.309754 & 3.00028 \\ 3.1 & 0.315873 & 3.10012 \end{array} \right)$$
Let $s$ be the length of the sagittus, $a$ the length of the arc, $r$ the radius, and $\theta$ the central angle. We are given $a$ and $s$. We know $a=r\theta$ and $ s = r-r\cos\left(\frac{\theta}{2}\right)$ so that $$ \frac{a}{s}=\frac{\theta}{1-\cos\left(\frac{\theta}{2}\right)}$$ A quick plot of $f(\theta)= \theta/(1-\cos(\theta/2))$ indicates that $f$ is one-to-one on $(0,\pi)$ so yes, $s$ and $a$ determine the angle. I don't think there is a simple closed-form formula for $\theta,$ however. You probably have to compute it numerically.
EDIT
At the OP's request, I'm adding some comments on how to compute this numerically. First, I think it's a little more convenient to consider the reciprocal of the expression I had before. $$ \frac{s}{a}=\frac{1-\cos\left(\frac{\theta}{2}\right)}{\theta}$$ Let $x=s/a,$ so that we want to solve $$f(\theta)=x\theta + \cos\left(\frac{\theta}{2}\right)-1=0$$ for $\theta,$ with $x$ given. The secant method is probably a good way to do this. You need a couple of starting values for $\theta_0$ and $\theta_1$ and you can use $1$ and $2$, I should think.
Let me know if this doesn't work well for you.