1. **State the problem:** We have an isosceles triangle with one side length 6.2 mm and an angle of 58° between that side and the base. We need to find the length $w$ of the base. 2. **Identify the triangle properties:** In an isosceles triangle, two sides are equal. Here, the side of length 6.2 mm is one of the equal sides. The angle given (58°) is between this side and the base $w$. 3. **Use the Law of Cosines:** To find the base $w$, we use the Law of Cosines formula: $$w^2 = a^2 + b^2 - 2ab \cos(C)$$ where $a$ and $b$ are the equal sides (both 6.2 mm), and $C$ is the angle between them (58°). 4. **Substitute values:** $$w^2 = 6.2^2 + 6.2^2 - 2 \times 6.2 \times 6.2 \times \cos(58^\circ)$$ 5. **Calculate each term:** $$6.2^2 = 38.44$$ $$2 \times 6.2 \times 6.2 = 76.88$$ $$\cos(58^\circ) \approx 0.5299$$ 6. **Calculate the product:** $$76.88 \times 0.5299 \approx 40.73$$ 7. **Calculate $w^2$:** $$w^2 = 38.44 + 38.44 - 40.73 = 76.88 - 40.73 = 36.15$$ 8. **Find $w$ by taking the square root:** $$w = \sqrt{36.15} \approx 6.0$$ 9. **Final answer:** The length $w$ is approximately 6.0 mm to 1 decimal place.