5. Points $A, B$ and $C$ divide the circle into three circular arcs $\overparen{A B}, \overparen{B C}$ and $\overparen{A C}$ whose lengths are in the ratio $3: 4: 5$. What are the measures of the interior angles of the triangle $A B C$ determined by these points?

## Problems worth 10 points:
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_46538cfe21908c5128afg-4.jpg?height=443&width=446&top_left_y=301&top_left_x=385)

Let the lengths of the circular arcs $\overparen{A B}, \overparen{B C}$, and $\overparen{A C}$ be denoted by $l_{1}, l_{2}$, and $l_{3}$, respectively, and their corresponding central angles by $\gamma_{1}, \alpha_{1}$, and $\beta_{1}$.

Since $l_{1}=\frac{3}{12} \cdot 2 r \pi=\frac{1}{4} \cdot 2 r \pi$ and $l_{1}=\frac{r \pi}{180^{\circ}} \cdot \gamma_{1}$, it follows that $\gamma_{1}=\frac{r \pi}{2} \cdot \frac{180^{\circ}}{r \pi}=90^{\circ}$.

Similarly, we obtain $\alpha_{1}=120^{\circ}$ and $\beta_{1}=150^{\circ}$.

The angles of the triangle $\alpha, \beta$, and $\gamma$ are the inscribed angles corresponding to the central angles $\alpha_{1}, \beta_{1}$, and $\gamma_{1}$, so finally $\alpha=60^{\circ}, \beta=75^{\circ}$, and $\gamma=45^{\circ}$.