Given the chord, & arc length of circular segment, how to define a formula for the height that is correct when chord length equals the arc length? 
In a circular segment, can we depend on the chord length, arch length, and add to it the angle between the tangent at the arc endpoints and the chord (if at all needed), to define the height of the segment.
We want the height to equal zero and the formula is still defined for chord length equal to arc length (and the angle between the tangent and chord is zero).
In other words, we need to deliberately not use radius, arc angle, or divide by the height.

The reason being that I'm using the problem solution in a computer software, and I can not deal with infinitely large radii, or even too big radii in float arithmetics. And intuitively it seems totally possible to depend on tangent angle with very small or zero value, to define a height of zero whenever encountered.
 A: The answer to Calculating the height of a circular segment at all points provided only chord and arc lengths shows that you have to solve the equation (where $\phi = \theta /2\in [0,\pi/2]$)
$$\phi= \frac{s}{a} \sin \phi$$
which in general can only be done numerically. If you have determined $\phi$, then
$$h = \frac{a}{2 \sin \phi} ( 1 - \cos \phi) = \frac{a \sin\phi}{2(1+ \cos \phi)} $$
since $\frac{1 - \cos \phi}{ \sin \phi}  = \frac{\sin \phi}{1 + \cos \phi}$. This applies also for $\phi = 0$ which is the "limit angle" for $s = a$. It gives $h = 0$ as desired.
Note that $\phi= \frac{s}{a} \sin \phi$ is equivalent to
$$\frac{a}{s} = \frac{\sin \phi}{\phi} = \sum_{n=1}^\infty (-1)^n\frac{\phi^{2n}} {(2n+1)!} = 1 - \frac{\phi^{2}}{3!} + \frac{\phi^{4}}{5!} - \ldots$$
which allows to get suitable numerical approximations.
By elementary geometry you see that the angle between the tangent at the arc endpoints and the chord is nothing else than $\phi$ (draw a picture). Thus, if you know this angle, you are done.
Remark: If you know $\phi$, then you do not need to calculate $\sin \phi$ and $\cos \phi$ by using the sine and cosine functions. In fact, you have
$$\sin \phi =  \frac{a}{s} \phi \\ \cos \phi =  \sqrt{1 - \frac{a^2}{s^2} \phi^2} \\ h = \frac{a^2 \phi}{2(s + \sqrt{s^2 - a^2\phi^2})}$$
