1. **State the problem:** We are given a triangle ABC with sides AC = $7\sqrt{3}$ cm, AB = $\sqrt{6}$ cm, and BC = $a$ cm. The angle at vertex A is 45°. We need to show that $a = \sqrt{n}$ for some integer $n$ and find the value of $n$. 2. **Use the Law of Cosines:** The Law of Cosines relates the sides and the included angle in a triangle: $$a^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle A)$$ 3. **Substitute the known values:** $$a^2 = (\sqrt{6})^2 + (7\sqrt{3})^2 - 2 \cdot \sqrt{6} \cdot 7\sqrt{3} \cdot \cos(45^\circ)$$ 4. **Simplify each term:** - $(\sqrt{6})^2 = 6$ - $(7\sqrt{3})^2 = 7^2 \cdot (\sqrt{3})^2 = 49 \cdot 3 = 147$ - $\cos(45^\circ) = \frac{\sqrt{2}}{2}$ So, $$a^2 = 6 + 147 - 2 \cdot \sqrt{6} \cdot 7\sqrt{3} \cdot \frac{\sqrt{2}}{2}$$ 5. **Simplify the product inside the subtraction:** $$2 \cdot \sqrt{6} \cdot 7\sqrt{3} \cdot \frac{\sqrt{2}}{2} = \cancel{2} \cdot \sqrt{6} \cdot 7\sqrt{3} \cdot \frac{\sqrt{2}}{\cancel{2}} = 7 \cdot \sqrt{6} \cdot \sqrt{3} \cdot \sqrt{2}$$ 6. **Multiply the square roots:** $$\sqrt{6} \cdot \sqrt{3} \cdot \sqrt{2} = \sqrt{6 \times 3 \times 2} = \sqrt{36} = 6$$ 7. **Calculate the subtraction term:** $$7 \times 6 = 42$$ 8. **Put it all together:** $$a^2 = 6 + 147 - 42 = 153 - 42 = 111$$ 9. **Express $a$ in terms of $n$:** $$a = \sqrt{111}$$ 10. **Conclusion:** We have shown that $a = \sqrt{n}$ where $n = 111$, an integer. **Final answer:** $a = \sqrt{111}$