How to calculate the radius of a Arc Segment given only the Arc Length and the Height of the arc segment? Most of the calculations I know and can find can solve for the radius if the cord length is known or the angle is known along with the height of the arc segment.  However I cannot find, nor figure out how to calculate the radius if only the arc segment and the height of the segment is known.
I have looked up several different methods on this and other site and the following are similar but not exactly what I am looking for.


*

*Calculate the radius of a circle given the chord length and height of a segment

*Calculating the height of a circular segment at all points provided only chord and arc lengths)


In CAD I can draw a arc like what is shown at (http://mathworld.wolfram.com/CircularSegment.html).  If I fix the mid point at (0,0), the set the cord horizontal, then add a dimension the segment height (i.e. h=2") and then dimension the arc length (i.e. s=10"), the geometry is fully constrained.  So I know there is a way to solve this.   
If I reversed the inputs and had the arc length (s) known and the radius (R) known I can calculate the segment height but I am having trouble reversing the equation.
 A: As Aretino answered, you  need some numerical methods to solve the equation $$L\big(1-\cos(\theta)\big)=2H\theta$$ (remember that there is no analytical solution to $x=\cos(x)$).
However, for $-\frac \pi 2 \leq \theta\leq\frac \pi 2$, you could use as a quite good approximation (see here) $$\cos(\theta) \simeq\frac{\pi ^2-4\theta^2}{\pi ^2+\theta^2}$$ which makes the equation $$2 H \theta^3-5 L \theta^2+2 \pi ^2 H \theta=0$$ which reduces to a quadratic $$2 H \theta^2-5 L \theta+2 \pi ^2 H =0$$ the solution of which providing a reasonable starting point for the iterative process.
For illustration purposes, let me consider $L=10$, $H=2$. The retained solution of the quadratic is $$\theta=\frac{1}{4} \left(25-\sqrt{625-16 \pi ^2}\right)\approx 0.846955$$ while the "exact" solution is $\approx 0.849952$.
Edit
Setting $\lambda=\frac{2H}L$, the initial equation write $$1-\cos(\theta)=\lambda\theta$$ Assuming $-\frac \pi 2 \leq \theta\leq\frac \pi 2$, the approximation leads to $$\theta_0=\frac{5-\sqrt{25-4 \pi ^2 \lambda ^2}}{2 \lambda }$$ Now, peform one iteration of Newton method using the exact equation; this will lead to $$\theta_1=\frac{\theta_0  \lambda +\cos (\theta_0 )-1}{\sin (\theta_0 )-\lambda }+\theta_0$$ For illustration, using $\lambda=\frac 25$ as before, this leads to $$\theta_0\approx 0.846955$$ $$\theta_1\approx 0.849961$$ while the exact solution woudld be $\approx 0.849952$. Continuing the iterations for many more figures, the itrates would be 
$$\left(
\begin{array}{cc}
 n & \theta_n \\
 0 & 0.846954969749869 \\
 1 & 0.849960525069669 \\
 2 & 0.849952018812858 \\
 3 & 0.849952018744878 
\end{array}
\right)$$ which is the solution for fifteen significant figures.
A: If $2\theta$ is the angle related to the given arc, having length $l$, radius $r$ and height $h$, then you have $l=2\theta r$ and $h=r(1-\cos\theta)$. By eliminating $r$ you are left with $l(1-\cos\theta)=2h\theta$, which in general can be solved for $\theta$ only numerically.
