1. **State the problem:** We have a scalene triangle ABC with sides BC = 85.94, AC = 74, and an angle of 45 degrees at vertex B. We want to find the length of side AB and the height from vertex A to base BC. 2. **Identify known values:** - Side BC = 85.94 - Side AC = 74 - Angle at B = 45 degrees 3. **Use the Law of Cosines to find AB:** The Law of Cosines states: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle B)$$ Substitute the values: $$AB^2 = 74^2 + 85.94^2 - 2 \times 74 \times 85.94 \times \cos(45^\circ)$$ Calculate each term: $$74^2 = 5476$$ $$85.94^2 \approx 7385.2836$$ $$\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$$ Calculate the product: $$2 \times 74 \times 85.94 \times 0.7071 \approx 2 \times 74 \times 85.94 \times 0.7071 = 2 \times 74 \times 60.75 \approx 2 \times 4495.5 = 8991$$ More precisely: $$2 \times 74 \times 85.94 \times 0.7071 = 2 \times 74 \times 85.94 \times 0.7071 \approx 8991.1$$ Now compute: $$AB^2 = 5476 + 7385.2836 - 8991.1 = 12861.2836 - 8991.1 = 3870.1836$$ Take the square root: $$AB = \sqrt{3870.1836} \approx 62.23$$ 4. **Find the height from A to BC:** The height $h$ can be found using the formula: $$h = AC \times \sin(\angle B)$$ Since $\sin(45^\circ) = 0.7071$, $$h = 74 \times 0.7071 \approx 52.13$$ **Final answers:** - Side AB $\approx 62.23$ - Height from A to BC $\approx 52.13$