1. **State the problem:** We need to find the values of $\cos x$ and $\cos z$ in a triangle with sides 7 m, 5 m, and 3 m, where angle $x$ is opposite the side of length 3 m, and angle $z$ is opposite the side of length 7 m. 2. **Recall the Law of Cosines:** For any triangle with sides $a$, $b$, and $c$, and angle $C$ opposite side $c$, the law states: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ 3. **Identify sides for angle $x$:** - Opposite side to $x$ is 3 m ($c=3$) - Other two sides are 7 m and 5 m ($a=7$, $b=5$) 4. **Calculate $\cos x$ using the Law of Cosines:** $$\cos x = \frac{7^2 + 5^2 - 3^2}{2 \times 7 \times 5} = \frac{49 + 25 - 9}{70} = \frac{65}{70}$$ 5. **Simplify the fraction for $\cos x$:** $$\frac{65}{70} = \frac{\cancel{65}^{13}}{\cancel{70}^{14}} = \frac{13}{14}$$ 6. **Identify sides for angle $z$:** - Opposite side to $z$ is 7 m ($c=7$) - Other two sides are 5 m and 3 m ($a=5$, $b=3$) 7. **Calculate $\cos z$ using the Law of Cosines:** $$\cos z = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30}$$ 8. **Simplify the fraction for $\cos z$:** $$\frac{-15}{30} = \frac{\cancel{-15}^{-1} }{\cancel{30}^{2}} = -\frac{1}{2}$$ **Final answers:** - $\cos x = \frac{13}{14}$ - $\cos z = -\frac{1}{2}$