5.59 Find the angles $\alpha, \beta$ and $\gamma$ of the first quadrant, if it is known that they form an arithmetic progression with a common difference of $\frac{\pi}{12}$, and their tangents form a geometric progression.
5.59 Let $\alpha=\beta-\frac{\pi}{12}, \gamma=\beta+\frac{\pi}{12}$. Since by condition $\operatorname{tg} \beta=$

$=q \operatorname{tg} \alpha, \operatorname{tg} \gamma=q^{2} \operatorname{tg} \alpha=q \operatorname{tg} \beta$, then

$\frac{\operatorname{tg} \beta}{\operatorname{tg}\left(\beta-\frac{\pi}{12}\right)}=\frac{\operatorname{tg}\left(\beta+\frac{\pi}{12}\right)}{\operatorname{tg} \beta}$, or $\operatorname{tg}^{2} \beta=\operatorname{tg}\left(\beta-\frac{\pi}{12}\right) \operatorname{tg}\left(\beta+\frac{\pi}{12}\right)$.

Transform the right side of this equation:

$\frac{\operatorname{tg} \beta-\operatorname{tg} \frac{\pi}{12}}{1+\operatorname{tg} \frac{\pi}{12} \operatorname{tg} \beta} \cdot \frac{\operatorname{tg} \beta+\operatorname{tg} \frac{\pi}{12}}{1-\operatorname{tg} \frac{\pi}{12} \operatorname{tg} \beta}=\frac{\operatorname{tg}^{2} \beta-\operatorname{tg}^{2} \frac{\pi}{12}}{1-\operatorname{tg}^{2} \frac{\pi}{12} \operatorname{tg}^{2} \beta}$

Then we get

$\operatorname{tg}^{2} \beta-\operatorname{tg}^{2} \frac{\pi}{12} \cdot \operatorname{tg}^{4} \beta=\operatorname{tg}^{2} \beta-\operatorname{tg}^{2} \frac{\pi}{12} ; \operatorname{tg}^{4} \beta=1 ;$

since $0<\beta<\frac{\pi}{2}$, we have $\operatorname{tg} \beta=1$, i.e., $\beta=\frac{\pi}{4}$.

Answer: $\alpha=\frac{\pi}{6} \beta=\frac{\pi}{4}, \gamma=\frac{\pi}{3}$.