How can I caculate the segment length necessary to separate a certain area from a sector under a given angle?

Question

Consider the geometry in this drawing.

A circular sector with radius $r$ is divided into two parts by a segment $l$ starting from one of the ends if its arc and meeting the opposite side of the sector at an angle $\alpha$. The sector area outside of $l$ (as indicated by blue shading in the drawing) is $A$.

The values $r$, $\alpha$ and $A$ are given, I need to find $l$. I also need the solution to be easily calculable (using a computer).

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With $\beta$ and $d$ as indicated in the drawing, one easily sees $$ \begin{gather} A = \frac{r^2}{2} \cdot (\beta - \sin\beta \cos\beta) + \frac{l^2}{2} \cdot \sin\alpha \cos\alpha \tag{1}\\ d = r \cdot \sin\beta = l \cdot \sin\alpha \tag{2}. \end{gather} $$ However, I can't seem to solve this for $l$. How can it be done? Is there maybe a better approach than the two equations above?

Answer 1

It looks to me that $$A=\frac{r^2 \beta}{2}-\frac{rl}{2}\sin(\alpha-\beta)$$ thus $$l=\frac{r^2\beta-2A}{r\sin(\alpha-\beta)}$$

Answer 2

We have,

$$\sin \beta = \frac dr = \frac{l\sin \alpha}{r} \\ \implies \beta = \sin^{-1} \left(\frac{l\sin \alpha}{r} \right)$$ Also, the area of the white triangle is $\frac 12 rl \sin (\alpha-\beta)$ .

$$\therefore A = \frac{r^2}{2} \beta - \frac 12 rl \sin(\alpha -\beta)$$ $$A=\frac{r^2}{2} \sin^{-1} \left( \frac{l\sin \alpha}{r} \right) - \frac 12 rl \sin \left(\alpha - \sin^{-1} \left(\frac{l\sin\alpha}{r} \right) \right)$$

I don’t see a more calculable approach, hopefully your computer can handle this.