1. **Problem statement:** We have an isosceles triangle with two equal sides each measuring 6.2 mm and an angle of 58° opposite the side of length $w$. We need to find the length $w$ to 1 decimal place. 2. **Formula used:** In any triangle, the Law of Cosines relates the sides and angles: $$w^2 = a^2 + b^2 - 2ab \cos(C)$$ where $a$ and $b$ are the equal sides (6.2 mm each), and $C$ is the angle opposite side $w$ (58°). 3. **Apply the Law of Cosines:** $$w^2 = 6.2^2 + 6.2^2 - 2 \times 6.2 \times 6.2 \times \cos(58^\circ)$$ 4. **Calculate each term:** $$6.2^2 = 38.44$$ $$2 \times 6.2 \times 6.2 = 76.88$$ $$\cos(58^\circ) \approx 0.5299$$ 5. **Substitute values:** $$w^2 = 38.44 + 38.44 - 76.88 \times 0.5299 = 76.88 - 40.74 = 36.14$$ 6. **Find $w$ by taking the square root:** $$w = \sqrt{36.14} \approx 6.0$$ 7. **Final answer:** The length $w$ is approximately **6.0 mm** to 1 decimal place.