1. Problem: We are given a quadrilateral ABCD with sides AB = 11 cm, BC = 4 cm, AC = 13 cm, CD = 8 cm, and angle \(\angle ADC = 80^\circ\). We need to find the size of the angle \(\angle ACB\). 2. To find \(\angle ACB\), consider triangle ABC with sides \(AB = 11\) cm, \(BC = 4\) cm, and \(AC = 13\) cm. 3. Use the Law of Cosines on triangle ABC to find \(\angle ACB\): $$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle ACB)$$ Substitute values: $$4^2 = 11^2 + 13^2 - 2 \cdot 11 \cdot 13 \cdot \cos(\angle ACB)$$ 4. Calculate the squares: $$16 = 121 + 169 - 286 \cdot \cos(\angle ACB)$$ $$16 = 290 - 286 \cdot \cos(\angle ACB)$$ 5. Rearrange to solve for \(\cos(\angle ACB)\): $$286 \cdot \cos(\angle ACB) = 290 - 16 = 274$$ $$\cos(\angle ACB) = \frac{274}{286} \approx 0.95804$$ 6. Find \(\angle ACB\) using the inverse cosine: $$\angle ACB = \cos^{-1}(0.95804) \approx 16.45^\circ$$ 7. Final answer: The size of the angle \(\angle ACB\) is approximately $16.45^\circ$.