Given the length $B$, horizontal width $C$ and the radius $R$, I need to calculate the start $(A)$ -/end $(A+\theta)$ angles.
So far I've got
$C=R(\cos(A)-\cos(A+\theta))$
where $\theta=B/R$
If I'm not mistaken, I need to isolate $A$ out in the form of $A=...$, which I am unable to do.
How do I isolate A? Am I thinking right?
This requires the sums to products formula.
$$\cos\theta-\cos\varphi=-2sin\left(\frac{\theta+\varphi}{2}\right)\sin\left(\frac{\theta-\varphi}{2}\right)$$
Applying it to your formula you get:
$$C=R(\cos(A)-\cos(A+\theta))$$
$$\frac{C}{R}=-2\sin\left(\frac{A+A+\theta}{2}\right)\sin\left(\frac{A-(A+\theta)}{2}\right)$$
$$\frac{C}{R}=-2\sin\left(\frac{2A+\theta}{2}\right)\sin\left(\frac{-\theta}{2}\right)$$
$$2\sin\left(\frac{2A+\theta}{2}\right)=\frac{C}{R\cdot\sin\left(\frac{\theta}{2}\right)}$$
$$\sin\left(\frac{2A+\theta}{2}\right)=\frac{C}{2R\cdot\sin\left(\frac{\theta}{2}\right)}$$
$$\frac{2A+\theta}{2}=\arcsin\left(\frac{C}{2R\cdot\sin\left(\frac{\theta}{2}\right)}\right)$$
$$A=\arcsin\left(\frac{C}{2R\cdot\sin\left(\frac{\theta}{2}\right)}\right)-\frac{\theta}{2}$$