1. **State the problem:** We need to find the angle $\theta$ at vertex B in the first triangle using the Law of Cosines. The sides are given as follows: side opposite $\theta$ is 42.75 units, and the two adjacent sides are 64.01 units and 35.51 units. 2. **Law of Cosines formula:** For a triangle with sides $a$, $b$, and $c$, and angle $\theta$ opposite side $a$, the Law of Cosines states: $$\cos(\theta) = \frac{b^2 + c^2 - a^2}{2bc}$$ 3. **Assign sides:** Let $a = 42.75$, $b = 64.01$, and $c = 35.51$. 4. **Calculate $\cos(\theta)$:** $$\cos(\theta) = \frac{64.01^2 + 35.51^2 - 42.75^2}{2 \times 64.01 \times 35.51}$$ Calculate each term: $$64.01^2 = 4097.28$$ $$35.51^2 = 1260.92$$ $$42.75^2 = 1827.56$$ So, $$\cos(\theta) = \frac{4097.28 + 1260.92 - 1827.56}{2 \times 64.01 \times 35.51} = \frac{3530.64}{4545.12} \approx 0.7769$$ 5. **Find $\theta$:** $$\theta = \cos^{-1}(0.7769) \approx 39.1^\circ$$ --- 6. **Second triangle problem:** Given side AC = 11 units and angle $\theta$ at vertex A, but no other data is provided, so we cannot solve for $\theta$ here without additional information. **Final answer:** The angle $\theta$ at vertex B in the first triangle is approximately $39.1^\circ$.