1. **Problem statement:** Determine the length of side $c$ in each triangle $\triangle ABC$ given the sides and angle. 2. **Formula used:** We use the Law of Cosines: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $a$ and $b$ are the sides enclosing angle $C$, and $c$ is the side opposite angle $C$. --- ### a) Given: - $BC = 41$ cm - $AC = 26$ cm - $\angle C = 30^\circ$ - Find $c = AB$ 3. Apply Law of Cosines: $$c^2 = 41^2 + 26^2 - 2 \times 41 \times 26 \times \cos(30^\circ)$$ 4. Calculate each term: $$41^2 = 1681$$ $$26^2 = 676$$ $$2 \times 41 \times 26 = 2132$$ $$\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$$ 5. Substitute values: $$c^2 = 1681 + 676 - 2132 \times 0.8660 = 2357 - 1847.512 = 509.488$$ 6. Find $c$: $$c = \sqrt{509.488} \approx 22.57$$ 7. Round to nearest tenth: $$c \approx 22.6 \text{ cm}$$ --- ### b) Given: - $AC = 6$ m - $AB = 10$ m - $\angle C = 45^\circ$ - Find $c = BC$ 8. Apply Law of Cosines: $$c^2 = 6^2 + 10^2 - 2 \times 6 \times 10 \times \cos(45^\circ)$$ 9. Calculate each term: $$6^2 = 36$$ $$10^2 = 100$$ $$2 \times 6 \times 10 = 120$$ $$\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$$ 10. Substitute values: $$c^2 = 36 + 100 - 120 \times 0.7071 = 136 - 84.852 = 51.148$$ 11. Find $c$: $$c = \sqrt{51.148} \approx 7.15$$ 12. Round to nearest tenth: $$c \approx 7.2 \text{ m}$$