The given equation is equivalent to $\lg \frac{a-b}{2} = \lg \sqrt{\frac{ab}{2}}$. Therefore
$$\frac{a-b}{2} = \sqrt{\frac{ab}{2}} \Leftrightarrow \frac{a^2-2ab+b^2}{4} = \frac{ab}{2} \Leftrightarrow a^2-4ab+b^2=0.$$
Since $a > b > 0$, by dividing with $b^2$, we have $(\frac{a}{b})^2 - 4\frac{a}{b} + 1 = 0$, with the roots $\frac{a}{b} = 2\pm\sqrt{3}$. Because of $\frac{a}{b} > 1$ we have $\frac{a}{b} = 2+\sqrt{3}$. On another side $\frac{a}{b} = \operatorname{tg}\alpha$.
From $\sin 2\alpha = \frac{2\operatorname{tg}\alpha}{1+\operatorname{tg}^2\alpha} = \frac{2(2+\sqrt{3})}{1+(2+\sqrt{3})^2} = \frac{1}{2}$ we have the two solutions for the angle $\alpha$, $\alpha = 75^\circ$ or $\alpha = 15^\circ$. Since $a > b$ we have $\alpha > \beta$, i.e. $\alpha = 75^\circ$ and $\beta = 15^\circ$.