Calculate the angle when the area of section is given as the % of total area of the circle This is not a homework. Just a sudden mathematical spark of my brain prompted me to simply calculate this.

In the diagram above the area of the hatched section is 10% of total area if the circle. What the angle CAB should be to satisfy this condition. I need non calculus solution.
So far I have done this...

But my final equation cannot be simplified further, or I don't know how to proceed further. What I need is an equation in which the alpha will be the title, so that I can find this alpha for any given area of the hatched section A.
 A: Another approach. From Thales' theorem we know that $\beta=90°$ and $\alpha+\gamma=90°$ Thus $\gamma=90°-\alpha$. 
And the triangle $BOC$ is an isosceles triangle. Therefore the angle at $B$ of the  triangle $BOC$ is $90°-\alpha$ as well. Then the angle at $0$ of $BOC$ is 
$\delta=180°-2\cdot (90°-\alpha)$=$2\alpha$
Now you can use the well known formula for the area of a circular segment:
$A=\frac{r^2}{2}\cdot\left(\frac{\delta \cdot \pi}{180}-\sin (\delta) \right) $
Remark:
You can also use the identity $\boxed{\sin(2\cdot \alpha)=2\cdot \cos(\alpha)\cdot \sin(\alpha)}$. 
Your intermediate result  $ A=\frac{\alpha \cdot \pi}{180}\cdot r^2-r^2\cdot \cos(\alpha)\cdot \sin(\alpha)$ becomes
$$A=\frac{\alpha \cdot \pi}{180}\cdot r^2-r^2\cdot \frac12\cdot  \sin(2\alpha)$$
A: Let segment ABC be the given such that [segment ABC] = [yellow] + [orange] = $\dfrac {1}{10}\pi r^2$. At this stage, C is just a point on the minor arc AB.

Construction: Draw the sector OBC with its central angle = $\angle BOC = 36^0 = … = \dfrac {\pi^c}{5}$. Let $x^c$ be the central angle of sector OCA.
Then, [sector OBC] = [pink] + [orange] = $\dfrac {1}{2} r^2 \dfrac {\pi}{5}^c$.
In other words, [pink] = [yellow] and therefore, [pink] + [blue] = [yellow] + [blue].
That is, area of $\triangle OAB$ (edited) = area of sector OAC.
∴ $\dfrac {1}{2} r^2 \sin (x + \dfrac {\pi}{5}) = \dfrac {1}{2} r^2 x^c$.
After simplification, we have $x = \sin (x + \dfrac {\pi}{5})$, which is still in transcendental (but a bit simpler). From Wolframalpha, $x = 0.998435^c$
