1. **Problem statement:** An isosceles triangle has two sides of length 11 cm each, and its area is 40 cm². We need to find the sizes of the interior angles to the nearest degree. 2. **Known values:** - Two equal sides: $a = b = 11$ cm - Area: $A = 40$ cm² 3. **Formula for the area of a triangle using two sides and the included angle:** $$A = \frac{1}{2}ab\sin(C)$$ where $C$ is the angle between sides $a$ and $b$. 4. **Substitute known values:** $$40 = \frac{1}{2} \times 11 \times 11 \times \sin(C)$$ 5. **Simplify:** $$40 = \frac{121}{2} \sin(C)$$ 6. **Isolate $\sin(C)$:** $$\sin(C) = \frac{40 \times 2}{121} = \frac{80}{121}$$ 7. **Calculate $C$:** $$C = \arcsin\left(\frac{80}{121}\right) \approx \arcsin(0.6612) \approx 41.4^\circ$$ 8. **Since the triangle is isosceles, the other two angles are equal:** Let each be $A = B$. 9. **Sum of interior angles in a triangle:** $$A + B + C = 180^\circ$$ $$2A + 41.4 = 180$$ 10. **Solve for $A$:** $$2A = 180 - 41.4 = 138.6$$ $$A = \frac{138.6}{2} = 69.3^\circ$$ 11. **Round to nearest degree:** $$C \approx 41^\circ, \quad A = B \approx 69^\circ$$ **Final answer:** The interior angles are approximately $41^\circ$, $69^\circ$, and $69^\circ$.