11. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_11f35ff0646caec62551g-3.jpg?height=471&width=782&top_left_y=981&top_left_x=180)
Solution: From the condition, it follows that $c > b$. Let's find a point $D$ on the segment $AB$ such that $AC = AD$. Then the triangle $ACD$ is isosceles and $\angle ACD = \angle ADC = 90^\circ - \alpha / 2$. The angle $\angle ADC$ is the external angle of the triangle $CBD$. Therefore, $\angle BCD + \beta = \angle ADC = 90^\circ - \frac{\alpha}{2} = \alpha + \beta$. Hence, $\angle BCD = \alpha$, and the triangles $CDB$ and $ABC$ are similar. We have $\frac{BD}{BC} = \frac{BC}{AB}$ or $\frac{c - b}{a} = \frac{a}{c'}$, from which it follows that $a^2 + bc - c^2 = 0$. The quadratic equation $c^2 - 3c - 4 = 0$ has a unique positive root $c = 4$.

Answer: 4.