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.
 A: 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.
A: Your approach is reasonable. You have correctly expressed the ratio of the areas of the triangle and the incircle in terms of the side lengths of the triangle and the radius of the incircle.
To proceed, and find find the angles you can follow these steps:
Use the formula for the sine of the angle $\alpha$ in terms of the sides of the triangle:
$$\sin \alpha = \frac{b}{c}$$
Substitute this into the equation:
$$\frac{\frac{a}{c} + \frac{b}{c} + 1}{\frac{a}{c} + \frac{b}{c} - 1} = 2\sqrt{3} + 3$$
to obtain:
$$\frac{\frac{a}{c} + \sin \alpha + 1}{\frac{a}{c} + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Use the formula for the cosine of the angle $\alpha$ in terms of the sides of the triangle:
$$\cos \alpha = \frac{a}{c}$$
Substitute this into the equation above to obtain:
$$\frac{\cos \alpha + \sin \alpha + 1}{\cos \alpha + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Use the identity $\cos^2 \alpha + \sin^2 \alpha = 1$ to express $\cos \alpha$ in terms of $\sin \alpha$:
$$\cos \alpha = \pm \sqrt{1 - \sin^2 \alpha} = \pm \sqrt{1 - \frac{b^2}{c^2}}$$
Substitute this into the equation above to obtain:
$$\frac{\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha + 1}{\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Simplify this equation to obtain:
$$(\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha - 2\sqrt{3})(\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha + 2\sqrt{3}) = 1$$
Solve this quadratic equation for $\sin \alpha$ using the quadratic formula.
Once you have found $\sin \alpha$, use it to find $\cos \alpha$.
Use these values to find the angles of the triangle using the formulas:
$$\alpha = \sin^{-1} \left( \frac{b}{c} \right)$$
$$\beta = \cos^{-1} \left( \frac{a}{c} \right)$$
$$\gamma = 90^\circ - \alpha - \beta$$
Note that there are two solutions for $\sin \alpha$, corresponding to the two possible values of $\cos \alpha$. You should check both solutions to ensure that you have found all possible solutions for the angles of the triangle
