1. Stating the problem: We have a triangle ABC where the sides satisfy the relation $10a = 12c = 13b$. We need to find the smallest angle in the triangle. 2. Express the sides in terms of a common variable. Let $x = 10a = 12c = 13b$. Then: $$a = \frac{x}{10}, \quad c = \frac{x}{12}, \quad b = \frac{x}{13}$$ 3. Identify the smallest side to find the smallest angle. Since the sides are proportional to $\frac{x}{10}$, $\frac{x}{12}$, and $\frac{x}{13}$: $$\frac{x}{13} < \frac{x}{12} < \frac{x}{10}$$ Thus, $b$ is the smallest side. 4. By the triangle property, the smallest angle is opposite the smallest side. So the smallest angle is $B$ opposite side $b$. 5. Use the Law of Cosines to find angle $B$: $$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$ 6. Substitute values: $$a = \frac{x}{10}, c = \frac{x}{12}, b = \frac{x}{13}$$ Therefore: $$\cos B = \frac{\left(\frac{x}{10}\right)^2 + \left(\frac{x}{12}\right)^2 - \left(\frac{x}{13}\right)^2}{2 \times \frac{x}{10} \times \frac{x}{12}}$$ Simplifying $x^2$ terms cancel: $$\cos B = \frac{\frac{1}{100} + \frac{1}{144} - \frac{1}{169}}{2 \times \frac{1}{10} \times \frac{1}{12}}$$ Calculate numerator: $$\frac{1}{100} + \frac{1}{144} = \frac{36}{3600} + \frac{25}{3600} = \frac{61}{3600}, \quad \frac{61}{3600} - \frac{1}{169}$$ Find common denominator $3600 \times 169 = 608400$: $$\frac{61 \times 169}{608400} - \frac{3600}{608400} = \frac{10289 - 3600}{608400} = \frac{6689}{608400}$$ Denominator: $$2 \times \frac{1}{10} \times \frac{1}{12} = \frac{2}{120} = \frac{1}{60}$$ Thus: $$\cos B = \frac{6689/608400}{1/60} = \frac{6689}{608400} \times 60 = \frac{6689 \times 60}{608400} = \frac{401340}{608400} \approx 0.6597$$ 7. Finally, find angle $B$: $$B = \arccos(0.6597) \approx 48.8^\circ$$ 8. This is the smallest angle of the triangle. **Answer:** The smallest angle is approximately $48.8^\circ$.