1. **Problem statement:** Given triangle ABC with sides AC = 26 inches, BC = 20 inches, and angle ABC = 95°, find the angle $\theta$ at vertex C. 2. **Formula used:** Use the Law of Cosines to find the side AB first, then use the Law of Sines or Law of Cosines again to find angle $\theta$. Law of Cosines: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ where $C$ is the angle opposite side $c$. 3. **Step 1: Find side AB.** Here, side AB is opposite angle $B = 95^\circ$, sides AC = 26, BC = 20. $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(95^\circ)$$ Calculate: $$AB^2 = 26^2 + 20^2 - 2 \times 26 \times 20 \times \cos(95^\circ)$$ $$AB^2 = 676 + 400 - 1040 \times \cos(95^\circ)$$ Since $\cos(95^\circ) \approx -0.0872$, $$AB^2 = 1076 - 1040 \times (-0.0872) = 1076 + 90.688 = 1166.688$$ $$AB = \sqrt{1166.688} \approx 34.16$$ inches. 4. **Step 2: Find angle $\theta$ at vertex C using Law of Sines:** Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Here, side AB = 34.16 opposite angle $C = \theta$, side BC = 20 opposite angle $A$, and angle $B = 95^\circ$. We want $\theta = C$, so: $$\frac{AB}{\sin \theta} = \frac{BC}{\sin 95^\circ}$$ Rearranged: $$\sin \theta = \frac{AB \times \sin 95^\circ}{BC}$$ Calculate: $$\sin \theta = \frac{34.16 \times \sin 95^\circ}{20}$$ Since $\sin 95^\circ \approx 0.9962$, $$\sin \theta = \frac{34.16 \times 0.9962}{20} = \frac{34.03}{20} = 1.7015$$ Since $\sin \theta$ cannot be greater than 1, this means the triangle configuration is impossible or we must use Law of Cosines to find $\theta$ instead. 5. **Step 3: Use Law of Cosines to find angle $\theta$:** $$\cos \theta = \frac{BC^2 + AC^2 - AB^2}{2 \times BC \times AC}$$ Substitute values: $$\cos \theta = \frac{20^2 + 26^2 - 34.16^2}{2 \times 20 \times 26} = \frac{400 + 676 - 1166.688}{1040} = \frac{-90.688}{1040} = -0.0872$$ 6. **Step 4: Calculate $\theta$:** $$\theta = \cos^{-1}(-0.0872) \approx 95^\circ$$ **Final answer:** The angle $\theta$ at vertex C is approximately $95^\circ$.