1. In a right-angled triangle with legs $a$ and $b, a>b$, the following equality holds:

$$
\lg \frac{a-b}{2}=\frac{1}{2}(\lg a+\lg b-\lg 2)
$$

Find the angles of the triangle.
Solution. The given equation is equivalent to the equation

$$
\lg \frac{a-b}{2}=\lg \sqrt{\frac{a b}{2}}
$$

so therefore

$$
\frac{a-b}{2}=\sqrt{\frac{a b}{2}} \Leftrightarrow \frac{a^{2}-2 a b+b^{2}}{4}=\frac{a b}{2} \Leftrightarrow a^{2}-4 a b+b^{2}=0 .
$$

Clearly $a>b>0$ and if we divide by $b^{2}$ we get a quadratic equation in $\frac{a}{b}$ in the form $\left(\frac{a}{b}\right)^{2}-4 \frac{a}{b}+1=0$, whose solutions are $\frac{a}{b}=2 \pm \sqrt{3}$. But $\frac{a}{b}>1$, so $\frac{a}{b}=2+\sqrt{3}$. On the other hand $\frac{a}{b}=\operatorname{tg} \alpha$, so $\sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{2(2+\sqrt{3})}{1+(2+\sqrt{3})^{2}}=\frac{1}{2}$. There are two possible solutions for $\alpha$ as an angle in a right triangle, which are $\alpha=75^{\circ}$ or $\alpha=15^{\circ}$, but since $a>b$ it must also be $\alpha>\beta$, i.e., $\alpha=75^{\circ}$ and $\beta=15^{0}$.