2. Given is a right-angled triangle $\triangle A B C, \measuredangle C=90^{\circ}$, for whose sides $a=BC, b=CA$ and $c=AB$ the following holds:

$$
c^{2}(3 a+b)=4 a^{3}
$$

Determine the angles in $\triangle A B C$.
2. Let $\alpha$ be the angle at vertex $A$. Then we have $a=c \sin \alpha$ and $b=c \cos \alpha$, so the given equation reduces to $3 \sin \alpha + \cos \alpha = 4 \sin^3 \alpha$, or $3 \sin \alpha - 4 \sin^3 \alpha = -\cos \alpha$.

This is further equivalent to $\sin 3 \alpha = -\sin (90^\circ - \alpha)$. Since $\alpha$ is an acute angle, we have $0^\circ < 3 \alpha < 270^\circ$, so it can only be $3 \alpha = 180^\circ + (90^\circ - \alpha)$, from which we calculate $\alpha = \frac{270^\circ}{4} = 67.5^\circ$ and $\beta = 90^\circ - \alpha = 22.5^\circ$.