1. **Problem statement:** We are given a triangle ABC with sides AC = $7\sqrt{3}$ cm, AB = $\sqrt{6}$ cm, BC = $a$ cm, and angle $\angle A = 45^\circ$. We want to show that $a = \sqrt{n}$ for some integer $n$ and find the value of $n$. 2. **Formula used:** We use the Law of Cosines, which states: $$a^2 = b^2 + c^2 - 2bc \cos(\theta)$$ where $a$ is the side opposite angle $\theta$, and $b$, $c$ are the other two sides. 3. **Assign values:** Here, $a = BC$, $b = AC = 7\sqrt{3}$, $c = AB = \sqrt{6}$, and $\theta = 45^\circ$. 4. **Apply Law of Cosines:** $$a^2 = (7\sqrt{3})^2 + (\sqrt{6})^2 - 2 \times 7\sqrt{3} \times \sqrt{6} \times \cos 45^\circ$$ 5. **Calculate squares:** $$(7\sqrt{3})^2 = 7^2 \times 3 = 49 \times 3 = 147$$ $$(\sqrt{6})^2 = 6$$ 6. **Calculate cosine:** $$\cos 45^\circ = \frac{\sqrt{2}}{2}$$ 7. **Substitute values:** $$a^2 = 147 + 6 - 2 \times 7\sqrt{3} \times \sqrt{6} \times \frac{\sqrt{2}}{2}$$ 8. **Simplify the multiplication:** $$2 \times 7\sqrt{3} \times \sqrt{6} \times \frac{\sqrt{2}}{2} = \cancel{2} \times 7 \times \sqrt{3} \times \sqrt{6} \times \frac{\sqrt{2}}{\cancel{2}} = 7 \times \sqrt{3} \times \sqrt{6} \times \sqrt{2}$$ 9. **Simplify the radicals:** $$\sqrt{3} \times \sqrt{6} \times \sqrt{2} = \sqrt{3 \times 6 \times 2} = \sqrt{36} = 6$$ 10. **Calculate the product:** $$7 \times 6 = 42$$ 11. **Final expression for $a^2$:** $$a^2 = 147 + 6 - 42 = 153 - 42 = 111$$ 12. **Conclusion:** $$a = \sqrt{111}$$ Since 111 is an integer, we have shown that $a = \sqrt{n}$ where $n = 111$. **Final answer:** $a = \sqrt{111}$