1. **Stating the problem:** In triangle ABC, angle $\beta$ is one third of angle $\alpha$, and angle $\beta$ is 20° larger than angle $\gamma$. Find the sizes of the angles $\alpha$, $\beta$, and $\gamma$. 2. **Formula and rules:** The sum of angles in any triangle is always 180°: $$\alpha + \beta + \gamma = 180^\circ$$ Given: $$\beta = \frac{1}{3} \alpha$$ $$\beta = \gamma + 20^\circ$$ 3. **Substitute $\beta$ from the second equation into the first:** $$\frac{1}{3} \alpha = \gamma + 20$$ From this, express $\gamma$: $$\gamma = \frac{1}{3} \alpha - 20$$ 4. **Substitute $\beta$ and $\gamma$ into the triangle sum equation:** $$\alpha + \beta + \gamma = 180$$ $$\alpha + \frac{1}{3} \alpha + \left(\frac{1}{3} \alpha - 20\right) = 180$$ 5. **Simplify the left side:** $$\alpha + \frac{1}{3} \alpha + \frac{1}{3} \alpha - 20 = 180$$ $$\alpha + \frac{2}{3} \alpha - 20 = 180$$ $$\frac{3}{3} \alpha + \frac{2}{3} \alpha - 20 = 180$$ $$\frac{5}{3} \alpha - 20 = 180$$ 6. **Add 20 to both sides:** $$\frac{5}{3} \alpha = 200$$ 7. **Multiply both sides by the reciprocal $\frac{3}{5}$:** $$\alpha = 200 \times \frac{3}{5}$$ $$\alpha = 200 \times \cancel{\frac{3}{5}}$$ $$\alpha = 120$$ 8. **Calculate $\beta$ and $\gamma$:** $$\beta = \frac{1}{3} \times 120 = 40$$ $$\gamma = \beta - 20 = 40 - 20 = 20$$ 9. **Check sum:** $$120 + 40 + 20 = 180$$ **Final answer:** $$\alpha = 120^\circ, \quad \beta = 40^\circ, \quad \gamma = 20^\circ$$