1. **Problem statement:** Given trapezium AECB with AE \parallel BC, find the angles $\angle x$, $\angle y$, and $\angle z$. 2. **Key property:** In trapezium AECB, since AE is parallel to BC, alternate interior angles and corresponding angles formed by a transversal are equal. 3. **Step 1: Analyze angles at vertex B.** - Angles near B are 28°, 17°, and $\angle y$. - These three angles form a straight line, so their sum is 180°. $$\angle y + 28 + 17 = 180$$ 4. **Calculate $\angle y$: ** $$\angle y = 180 - 28 - 17 = 135$$ 5. **Step 2: Analyze angles at vertex C.** - Given $\angle C = 69^\circ$. 6. **Step 3: Use parallel lines property to find $\angle x$.** - Since AE \parallel BC and AB is a transversal, $\angle x$ and the 30° angle are alternate interior angles. - Therefore, $\angle x = 30^\circ$. 7. **Step 4: Find $\angle z$ at vertex E.** - Angles at E are $\angle z$ and 28° (given near B but related by parallel lines). - Since AE \parallel BC and CE is a transversal, $\angle z$ and 69° (at C) are supplementary. $$\angle z + 69 = 180$$ 8. **Calculate $\angle z$: ** $$\angle z = 180 - 69 = 111$$ **Final answers:** $$\angle x = 30^\circ, \quad \angle y = 135^\circ, \quad \angle z = 111^\circ$$