## Task B-2.2.

In a right-angled triangle where $c$ is the length of the hypotenuse, $a, b$ are the lengths of the legs, and $\alpha, \beta$ are the angles opposite to them respectively, the inequality $5 c^{4} \geqslant 6 a^{2} c^{2}+8 b^{4}$ holds. Determine all possible values that the angles $\alpha$ and $\beta$ can take in this triangle.
## Solution.

Apply one of the forms of the Pythagorean theorem to the inequality $5 c^{4} \geqslant 6 a^{2} c^{2}+8 b^{4}$, for example, $b^{2}=c^{2}-a^{2}$. By substitution, we get

$5 c^{4} \geqslant 6 a^{2} c^{2}+8\left(c^{2}-a^{2}\right)^{2}$.

After rearranging, we obtain $8 a^{4}-10 a^{2} c^{2}+3 c^{4} \leqslant 0$.

Divide the obtained inequality by $c^{4}$, and we get

$8\left(\frac{a}{c}\right)^{4}-10\left(\frac{a}{c}\right)^{2}+3 \leqslant 0$.

Since $\sin \alpha=\frac{a}{c}$, it follows that $8 \sin ^{4} \alpha-10 \sin ^{2} \alpha+3 \leqslant 0$.

After introducing the substitution $t=\sin ^{2} \alpha$, we get the inequality $8 t^{4}-10 t^{2}+3 \leqslant 0$.

The solution to this inequality is $\frac{1}{2} \leqslant t \leqslant \frac{3}{4}$. Therefore, $\frac{1}{2} \leqslant \sin ^{2} \alpha \leqslant \frac{3}{4}$.

Since $\sin \alpha \in (0,1]$, it follows that $\frac{\sqrt{2}}{2} \leqslant \sin \alpha \leqslant \frac{\sqrt{3}}{2}$, or $\alpha \in [45^{\circ}, 60^{\circ}]$.

Given $\beta=90^{\circ}-\alpha$, it follows that $\beta \in [30^{\circ}, 45^{\circ}]$.