1. **State the problem:** We need to find the values of angles $x$ and $y$ in the given geometric figure with quadrilateral ABCD, straight line CDE, and isosceles triangle AFB. 2. **Given information:** - Quadrilateral ABCD with angles at B = 130°, C = 70°, D = 104°, and angle at A split into $x^\circ$ and 30°. - CDE is a straight line, so angles on this line sum to 180°. - Triangle AFB is isosceles, so two of its angles are equal. 3. **Find $x$:** Since ABCD is a quadrilateral, the sum of its interior angles is 360°. $$x + 30 + 130 + 70 + 104 = 360$$ Combine known angles: $$30 + 130 + 70 + 104 = 334$$ So, $$x + 334 = 360$$ Subtract 334 from both sides: $$x = 360 - 334 = 26$$ 4. **Find $y$:** Triangle AFB is isosceles, so two angles are equal. The triangle's angles sum to 180°. The angles at A and B are $x + 30 = 26 + 30 = 56^6$ and 130°, but since 130° is at B in the quadrilateral, the triangle AFB's angles are at A, F, and B. We know angle at A is $x + 30 = 56^6$. Angle at B is 130° (given). Sum of angles in triangle AFB: $$56 + y + 130 = 180$$ Combine known angles: $$186 + y = 180$$ Subtract 186 from both sides: $$y = 180 - 186 = -6$$ A negative angle is impossible, so the isosceles triangle must have equal sides at A and F or B and F. Since AFB is isosceles, and angles at A and B are different, the equal angles must be at F and either A or B. Assuming angles at F and B are equal: Let $y = 130^6$ (angle at F), then angle at A is $x + 30 = 56^6$. Sum of angles in triangle AFB: $$56 + 130 + y = 180$$ But $y$ is 130°, so sum is: $$56 + 130 + 130 = 316$$ Too large. Assuming angles at F and A are equal: Let $y = 56^6$ (angle at F), angle at B is 130°. Sum: $$56 + 56 + 130 = 242$$ Too large. Try the base angles equal to $y$ and $x + 30$: Let angles at A and F be equal: $$y = x + 30 = 56$$ Sum of triangle angles: $$56 + 56 + 130 = 242$$ No. Try angles at F and B equal: $$y = 130$$ Sum: $$56 + 130 + 130 = 316$$ No. Try angles at A and B equal: $$56 = 130$$ No. Since the triangle is isosceles, two sides equal, so two angles equal. Given the above, the only possibility is that the base angles are $y$ and $x + 30$. Sum of triangle angles: $$y + y + 130 = 180$$ Because two equal angles $y$ and one angle 130°. Simplify: $$2y + 130 = 180$$ Subtract 130: $$2y = 50$$ Divide by 2: $$y = 25$$ 5. **Final answers:** $$x = 26^6$$ $$y = 25^6$$