1. **State the problem:** We are given a triangle ABC with sides AB = 4 cm, AC = 5 cm, and BC = 6 cm. We need to show that $\cos A = \frac{1}{8}$. 2. **Recall the Law of Cosines:** For any triangle with sides $a$, $b$, and $c$, and angle $A$ opposite side $a$, the law states: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ 3. **Assign sides relative to angle A:** Angle $A$ is opposite side $BC$, so $a = BC = 6$, $b = AB = 4$, and $c = AC = 5$. 4. **Substitute values into the formula:** $$\cos A = \frac{4^2 + 5^2 - 6^2}{2 \times 4 \times 5} = \frac{16 + 25 - 36}{40} = \frac{41 - 36}{40} = \frac{5}{40}$$ 5. **Simplify the fraction:** $$\frac{5}{40} = \frac{1}{8}$$ 6. **Conclusion:** We have shown that $\cos A = \frac{1}{8}$ as required.