1. **State the problem:** We are given a right-angled triangle ABC with right angle at B, sides AB = 4 m, BC = 6 m, and point D on BC such that BD = 1 m and DC = 5 m. We need to use the cosine rule to show that $\cos A + \cos C = \frac{7}{8}$ and that $\cos(A + C) = -\frac{9}{16}$. 2. **Recall the cosine rule:** For any triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively, the cosine rule states: $$c^2 = a^2 + b^2 - 2ab \cos C$$ Similarly for other angles. 3. **Identify sides and angles:** Since $\angle B = 90^\circ$, triangle ABC is right-angled at B. - $AB = 4$ (side adjacent to angle C) - $BC = 6$ (side adjacent to angle A) - $AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ (hypotenuse) 4. **Calculate $\cos A$ and $\cos C$ using definitions:** - $\cos A = \frac{\text{adjacent side to } A}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$ - $\cos C = \frac{AB}{AC} = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}}$ 5. **Sum $\cos A + \cos C$:** $$\cos A + \cos C = \frac{3}{\sqrt{13}} + \frac{2}{\sqrt{13}} = \frac{5}{\sqrt{13}}$$ 6. **Rationalize denominator:** $$\frac{5}{\sqrt{13}} = \frac{5}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{5\sqrt{13}}{13}$$ 7. **Evaluate $\frac{5\sqrt{13}}{13}$ numerically:** Since $\sqrt{13} \approx 3.605551275$, $$\frac{5 \times 3.605551275}{13} = \frac{18.027756375}{13} \approx 1.3867$$ This is not $\frac{7}{8} = 0.875$, so we must use the cosine rule instead. 8. **Use cosine rule to find $\cos A$ and $\cos C$:** - Opposite side to $A$ is $BC = 6$ - Opposite side to $C$ is $AB = 4$ - Side $a = BC = 6$, $b = AB = 4$, $c = AC = 2\sqrt{13}$ Using cosine rule for angle $A$: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$6^2 = 4^2 + (2\sqrt{13})^2 - 2 \times 4 \times 2\sqrt{13} \cos A$$ $$36 = 16 + 4 \times 13 - 16 \sqrt{13} \cos A$$ $$36 = 16 + 52 - 16 \sqrt{13} \cos A$$ $$36 = 68 - 16 \sqrt{13} \cos A$$ Rearranged: $$16 \sqrt{13} \cos A = 68 - 36 = 32$$ $$\cos A = \frac{32}{16 \sqrt{13}} = \frac{2}{\sqrt{13}}$$ Similarly for angle $C$: $$c^2 = a^2 + b^2 - 2ab \cos C$$ $$ (2\sqrt{13})^2 = 6^2 + 4^2 - 2 \times 6 \times 4 \cos C$$ $$52 = 36 + 16 - 48 \cos C$$ $$52 = 52 - 48 \cos C$$ $$48 \cos C = 52 - 52 = 0$$ $$\cos C = 0$$ This contradicts the previous step, so we must check carefully. 9. **Re-examine angle labels:** Since $B$ is right angle, $A$ and $C$ are acute angles. - Side opposite $A$ is $BC = 6$ - Side opposite $C$ is $AB = 4$ - Side opposite $B$ is $AC = 2\sqrt{13}$ Using cosine rule for $\cos A$: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$6^2 = 4^2 + (2\sqrt{13})^2 - 2 \times 4 \times 2\sqrt{13} \cos A$$ $$36 = 16 + 52 - 16 \sqrt{13} \cos A$$ $$16 \sqrt{13} \cos A = 68 - 36 = 32$$ $$\cos A = \frac{32}{16 \sqrt{13}} = \frac{2}{\sqrt{13}}$$ Similarly for $\cos C$: $$c^2 = a^2 + b^2 - 2ab \cos C$$ $$ (2\sqrt{13})^2 = 6^2 + 4^2 - 2 \times 6 \times 4 \cos C$$ $$52 = 36 + 16 - 48 \cos C$$ $$48 \cos C = 52 - 52 = 0$$ $$\cos C = 0$$ This suggests $\cos C = 0$, which is impossible for an acute angle. The error is in labeling sides for cosine rule. 10. **Correct labeling:** - Side $a$ opposite $A$ is $BC = 6$ - Side $b$ opposite $B$ is $AC = 2\sqrt{13}$ - Side $c$ opposite $C$ is $AB = 4$ Use cosine rule for $\cos A$: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$6^2 = (2\sqrt{13})^2 + 4^2 - 2 \times 2\sqrt{13} \times 4 \cos A$$ $$36 = 52 + 16 - 16 \sqrt{13} \cos A$$ $$16 \sqrt{13} \cos A = 68 - 36 = 32$$ $$\cos A = \frac{32}{16 \sqrt{13}} = \frac{2}{\sqrt{13}}$$ Use cosine rule for $\cos C$: $$c^2 = a^2 + b^2 - 2ab \cos C$$ $$4^2 = 6^2 + (2\sqrt{13})^2 - 2 \times 6 \times 2\sqrt{13} \cos C$$ $$16 = 36 + 52 - 24 \sqrt{13} \cos C$$ $$24 \sqrt{13} \cos C = 88 - 16 = 72$$ $$\cos C = \frac{72}{24 \sqrt{13}} = \frac{3}{\sqrt{13}}$$ 11. **Sum $\cos A + \cos C$:** $$\cos A + \cos C = \frac{2}{\sqrt{13}} + \frac{3}{\sqrt{13}} = \frac{5}{\sqrt{13}}$$ Rationalize: $$\frac{5}{\sqrt{13}} = \frac{5 \sqrt{13}}{13}$$ 12. **Evaluate $\frac{5 \sqrt{13}}{13}$ numerically:** $$\approx \frac{5 \times 3.605551275}{13} = \frac{18.027756375}{13} \approx 1.3867$$ This is not $\frac{7}{8} = 0.875$, so the problem likely assumes different side lengths or a different triangle. 13. **Show $\cos(A + C) = -\frac{9}{16}$:** Since $A + C = 90^\circ$ (right angle at B), $$\cos(A + C) = \cos 90^\circ = 0$$ This contradicts $-\frac{9}{16}$. 14. **Conclusion:** The given data and requested results do not match the right-angled triangle with sides 4 and 6. Possibly the problem refers to a different triangle or additional information is needed. **Final answer:** Using the cosine rule with the given triangle, we find $$\cos A + \cos C = \frac{5 \sqrt{13}}{13}$$ which is approximately 1.3867, not $\frac{7}{8}$. Also, $$\cos(A + C) = 0 \neq -\frac{9}{16}$$ Hence, the problem as stated cannot be shown with the given data.