1. **State the problem:** We need to find the size of angle $\theta$ at vertex B in a triangle with sides AB = 52 cm, BC = 43 cm, and AC = 77 cm. 2. **Formula used:** To find an angle when all three sides are known, we use the Law of Cosines: $$\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}$$ where $a$ and $b$ are the sides adjacent to angle $\theta$, and $c$ is the side opposite $\theta$. 3. **Identify sides:** Here, angle $\theta$ is at vertex B, so the sides adjacent to $\theta$ are AB = 52 cm and BC = 43 cm, and the side opposite $\theta$ is AC = 77 cm. 4. **Apply the formula:** $$\cos(\theta) = \frac{52^2 + 43^2 - 77^2}{2 \times 52 \times 43}$$ 5. **Calculate squares:** $$52^2 = 2704, \quad 43^2 = 1849, \quad 77^2 = 5929$$ 6. **Substitute values:** $$\cos(\theta) = \frac{2704 + 1849 - 5929}{2 \times 52 \times 43} = \frac{-1376}{4472}$$ 7. **Simplify:** $$\cos(\theta) \approx -0.3077$$ 8. **Find angle $\theta$:** $$\theta = \cos^{-1}(-0.3077)$$ 9. **Calculate using inverse cosine:** $$\theta \approx 108^\circ$$ **Final answer:** The size of angle $\theta$ is approximately **108 degrees**.