Finding the angles of a right triangle if $\frac{\text{area of triangle}}{\text{area of incircle}}=\frac{2\sqrt{3}+3}{\pi}$ 
Let $ABC$ be a right triangle with $\measuredangle ACB=90^\circ$. If $k(O;r)$ is the incircle of the triangle and
$$\dfrac{S_{ABC}}{S_k}=\dfrac{2\sqrt3+3}{\pi}$$ find the angles of the triangle.
($S$ denotes the areas of the triangle and the circle.)

The area of the triangle is $S=pr$, where $p$ is the semiperemeter, the area of the circle is $S_k=\pi r^2,$ so $$\dfrac{pr}{\pi r^2}=\dfrac{2\sqrt3+3}{\pi}\\\dfrac{p}{r}=2\sqrt3+3\\\dfrac{\frac{a+b+c}{2}}{\frac{a+b-c}{2}}=2\sqrt3+3\\\dfrac{\frac{a}{c}+\frac{b}{c}+1}{\frac{a}{c}+\frac{b}{c}-1}=2\sqrt3+3\\\ \dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}=2\sqrt3+3$$ I don't know if my approach is reasonable and don't see what can be done from here.

Rewrite $$\dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}$$
as
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1.$$
Then
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1=2\sqrt3+3\implies \dfrac{1}{\sin\alpha+\cos\alpha-1}=\sqrt3+1.$$
Rearranging, we get
$$\sin\alpha + \cos\alpha =\frac{1}{\sqrt{3}+1}+1=\frac{\sqrt{3}-1}{2}+1=\frac{\sqrt{3}}{2}+\frac{1}{2}.$$
We can see that $\alpha = 30^{\circ}, 60^{\circ}$ are solutions.
Using the identity $\sin\alpha + \cos\alpha = \sqrt{2}\sin(\alpha + 45^{\circ})$, we can see that these are also the only solutions.
