1. **Problem 6:** Use the Law of Cosines to find angle $\theta$ at vertex B in a triangle with sides $AB=42.75$, $BC=64.01$, and $AC=35.51$. 2. The Law of Cosines formula for angle $B$ is: $$\cos(\theta) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. Substitute the given side lengths: $$\cos(\theta) = \frac{42.75^2 + 64.01^2 - 35.51^2}{2 \cdot 42.75 \cdot 64.01}$$ 4. Calculate the squares: $$42.75^2 = 1827.56$$ $$64.01^2 = 4097.28$$ $$35.51^2 = 1260.92$$ 5. Substitute these values: $$\cos(\theta) = \frac{1827.56 + 4097.28 - 1260.92}{2 \cdot 42.75 \cdot 64.01} = \frac{4663.92}{5473.37}$$ 6. Calculate the denominator: $$2 \cdot 42.75 \cdot 64.01 = 5473.37$$ 7. Calculate the cosine value: $$\cos(\theta) = 0.8519$$ 8. Find the angle $\theta$ by taking the inverse cosine: $$\theta = \cos^{-1}(0.8519) = 31.5^\circ$$ (rounded to one decimal place) 9. **Problem 7:** Use the Law of Cosines to find angle $\theta$ at vertex C in a triangle where side $AC=11$. Other sides are not given explicitly, so we cannot solve this without additional information. **Final answers:** - For problem 6, $\theta = 31.5^\circ$ - For problem 7, insufficient data to determine $\theta$.