Problem 1. Determine the internal angles of a triangle if it is known that the size of one angle is $\frac{8}{15}$ of the size of another angle and $\frac{4}{11}$ of the size of the third angle.
Solution. Let $\alpha, \beta$ and $\gamma$ denote the interior angles of the triangle. From the condition of the problem, it follows that $\alpha=\frac{8}{15} \beta$ and $\alpha=\frac{4}{11} \gamma$. Then $\beta=\frac{15}{8} \alpha$ and $\gamma=\frac{11}{4} \alpha$. Since the sum of the interior angles of the triangle is $180^{\circ}$, we get $\alpha+\frac{15}{8} \alpha+\frac{11}{4} \alpha=180^{0}$. The solution to the equation is $\alpha=32^{0}$. Then $\beta=60^{\circ}$ and $\gamma=88^{0}$.