Arc length to different chord relation Figure 1: 
If I know $\theta$, $c$ and $h$, how can I find out $a$?
 A: In the green isoceles triangle,
$$\color{green}c=2\color{green}h\tan\frac\theta2$$
and the corresponding arc (no shown)
$$\color{green}a=\theta\frac{\color{green}h}{\cos\dfrac\theta2}.$$
Then by similarity,
$$\color{blue}a=\color{green}a\frac{\color{green}h+\color{red}h}{\color{green}h}=\theta\frac{\color{green}{\dfrac c2}\cot\dfrac\theta2+\color{red}h}{\cos\dfrac\theta2}.$$
A: In terms of the radius $r$ and the central angle $\theta$ (measured in radians), the length of the arc $a$ is 
$$a = r\theta $$
We only need to find the radius. 

I will assume that this is a circle sector, and that the segment with length $h$ is perpendiculat to the segment with length $c$. Then the problem should be solvable. 
First, label the points  in the circle sector as follows: 
Extend line segment $TU$ to $X$. 
Note that $XW = XV = r$, $TU = h$, $YT = TZ = c/2$, and $$\angle TXY = \angle TXZ = \theta /2$$
Also, $\angle XTY$ and $\angle XUW$ are both right angles. Triangles $\triangle  XTY$ and $\triangle XUW$ are therefore similar triangles.

Let $b$ be the length of segment $XT$, and let $d$ be the length of $XY$. Then by similarity:
$$\frac{r}{d} =\frac{b+h}{b}$$
Applying the law of sines to $\triangle XTY$, we get
$$\frac{\sin (\theta /2)}{c/2} = \frac{\sin 90^\circ}{d} = \frac{1}{d}$$
Solving for $d$ gives
$$d = \frac{c}{2\sin (\theta /2)}$$
Applying the law of sines to $\triangle XUW$, we get
$$\frac{\sin \left(90^\circ - (\theta /2)\right)}{b+h} = \frac{\sin 90^\circ}{r} = \frac{1}{r}$$
Solving for $b$ gives:
$$b = r\cos(\theta / 2) - h$$

Now, we can use the similarity relation we derived earlier, and plug in our equations for $b$ and $d$:
$$\frac{r}{d} = \frac{b+h}{b}$$
$$\frac{2r\sin(\theta / 2)}{c} = \frac{r\cos (\theta / 2)}{r\cos (\theta / 2) - h }$$
The $r$ factors cancel from the numerators:
$$\frac{2\sin(\theta / 2)}{c} = \frac{\cos (\theta / 2)}{r\cos (\theta / 2) - h }$$
Cross-multiplying and using the double angle sine identity gives:
$$r \sin \theta -  2h \sin (\theta / 2) =  c \cdot \cos (\theta / 2)$$
Finally, we solve for $r$:
$$r = \frac{2h \sin (\theta / 2) + c \cdot \cos (\theta / 2)}{\sin \theta}$$

We multiply by $\theta$ to get the final answer for the length of the arc $a$:
$$ a = \boxed{\frac{2h\theta \sin(\theta / 2) + c\theta \, \cos (\theta / 2)}{\sin \theta}}$$
