6. What are the acute angles of a right triangle if for its legs $a, b$ and hypotenuse (20) $c$ it holds that $4 a b=c^{2} \sqrt{3}$?
First solution. For the given right-angled triangle, we have $a^{2}+b^{2}=c^{2}$.

For the angle $\alpha$ opposite the side of length $a$, by substituting into the given equation, we get:

$$
\begin{aligned}
4 a b & =\sqrt{3} a^{2}+\sqrt{3} b^{2} \quad /: b^{2} \\
4 \frac{a}{b} & =\sqrt{3}\left(\frac{a}{b}\right)^{2}+\sqrt{3} \\
\sqrt{3} \operatorname{tg}^{2} \alpha-4 \operatorname{tg} \alpha+\sqrt{3} & =0
\end{aligned}
$$

(10 points)

From here, we get:

$\operatorname{tg} \alpha_{1}=\sqrt{3} \quad \mathrm{tj} . \quad \alpha_{1}=60^{\circ}, \beta_{1}=30^{\circ}$ or

$\operatorname{tg} \alpha_{2}=\frac{\sqrt{3}}{3} \quad$ i.e. $\quad \alpha_{2}=30^{\circ}, \beta_{2}=60^{\circ}$.

The acute angles of this triangle are $30^{\circ}$ and $60^{\circ}$.