1. **State the problem:** We have three regular nonagons (9-sided polygons) A, B, and C. Polygons A and B share a side. We need to find the sum of the angles $x$ and $y$ formed between polygons A and C, and B and C respectively. 2. **Find the interior angle of a regular nonagon:** The formula for the interior angle of a regular polygon with $n$ sides is $$\text{Interior angle} = \frac{(n-2) \times 180^\circ}{n}$$ For a nonagon, $n=9$, so $$\text{Interior angle} = \frac{(9-2) \times 180^\circ}{9} = \frac{7 \times 180^\circ}{9} = 140^\circ$$ 3. **Find the exterior angle of a regular nonagon:** The exterior angle is the supplement of the interior angle, $$\text{Exterior angle} = 180^\circ - 140^\circ = 40^\circ$$ 4. **Analyze the angles $x$ and $y$:** Since polygons A and B share a side, the angles $x$ and $y$ are formed outside the polygons at the vertices where polygon C meets polygons A and B. 5. **Sum of angles around point where polygons meet:** At the vertex where polygons A, B, and C meet, the sum of the exterior angles around that point is $360^\circ$. 6. **Calculate the sum $x + y$:** Each polygon contributes an exterior angle of $40^\circ$. Since polygons A and B share a side, the angles $x$ and $y$ are the exterior angles of polygons A and B adjacent to polygon C. Therefore, $$x + y = 40^\circ + 40^\circ = 80^\circ$$ **Final answer:** $$\boxed{80^\circ}$$