What are the acute angles $\alpha$ and $\beta$ if the following system of equations is true for them?

$$
\begin{aligned}
2 \sin 2 \beta & =3 \sin 2 \alpha \\
\operatorname{tg} \beta & =3 \operatorname{tg} \alpha
\end{aligned}
$$
Solution. Consider the triangle $ABC$ such that $\angle CAB = \alpha$ and $\angle CBA = \beta$. Draw the altitude from vertex $C$ to the line containing side $AB$, and let the foot of the altitude be $D$. Let $AD = a$ and $BD = b$.

$$
AC = \sqrt{a^2 + m^2}, \quad BC = \sqrt{b^2 + m^2}
$$

From the right triangles,

$$
\tan \alpha = \frac{m}{a} \quad \text{and} \quad \tan \beta = \frac{m}{b}
$$

According to equation (2), $\frac{m}{b} = 3 \cdot \frac{m}{a}$, and thus $a = 3b$. Furthermore,

$$
\sin \alpha = \frac{m}{\sqrt{a^2 + m^2}}, \quad \sin \beta = \frac{m}{\sqrt{b^2 + m^2}}, \quad \cos \alpha = \frac{a}{\sqrt{a^2 + m^2}}, \quad \cos \beta = \frac{b}{\sqrt{b^2 + m^2}}
$$

Using the double-angle identity (1) can be rewritten as: $2 \sin \beta \cos \beta = 3 \sin \alpha \cos \alpha$. Substituting the values of $\sin \alpha, \cos \alpha, \sin \beta, \cos \beta$ obtained earlier:

$$
2 \cdot \frac{m}{\sqrt{m^2 + b^2}} \cdot \frac{b}{\sqrt{m^2 + b^2}} = 3 \cdot \frac{m}{\sqrt{m^2 + a^2}} \cdot \frac{a}{\sqrt{m^2 + a^2}}
$$

Simplify by $m \neq 0$. Perform the indicated operations in the denominators, and substitute $a$ with the previously obtained $3b$:

$$
\frac{2b}{m^2 + b^2} = \frac{9b}{m^2 + 9b^2}
$$

Bring to a common denominator and rearrange the equation. The result is:

$$
7m^2 = 9b^2, \quad \text{hence} \quad m = \frac{3\sqrt{7}}{7}b
$$

Therefore,

$$
\begin{array}{rlrl}
\tan \alpha & = \frac{m}{a} & = \frac{3\sqrt{7}b}{7 \cdot 3b} = \frac{\sqrt{7}}{7}, & \text{so} \quad \alpha \approx 20.70^\circ \\
\text{and} \tan \beta & = \frac{m}{b} = \frac{3\sqrt{7}}{7}, & \text{so} \quad \beta \approx 48.59^\circ
\end{array}
$$