1. **Problem statement:** We are given a triangle ABC with sides AC = 130, side from C to the unlabeled vertex = 110, and an angle of 38° at the unlabeled vertex adjacent to the side of length 110. We need to find the angle $x$ at vertex C, which is opposite side AC. 2. **Identify known elements:** - Side $AC = 130$ - Side adjacent to angle 38° (let's call it $BC$) = 110 - Angle at the unlabeled vertex (let's call it $B$) = 38° - Angle at vertex $C$ is $x$ (unknown) 3. **Use the Law of Sines:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a,b,c$ are sides opposite angles $A,B,C$ respectively. 4. **Assign sides and angles:** - Side opposite angle $C$ is $AB$ (unknown length) - Side opposite angle $B$ is $AC = 130$ - Side opposite angle $A$ is $BC = 110$ Since we know side $AC=130$ opposite angle $B=38^\circ$, and side $BC=110$ opposite angle $A$, we can find angle $A$ first: $$\frac{110}{\sin A} = \frac{130}{\sin 38^\circ}$$ 5. **Calculate $\sin A$:** $$\sin A = \frac{110 \times \sin 38^\circ}{130}$$ Calculate $\sin 38^\circ \approx 0.6157$: $$\sin A = \frac{110 \times 0.6157}{130} = \frac{67.727}{130} \approx 0.5217$$ 6. **Find angle $A$:** $$A = \arcsin(0.5217) \approx 31.4^\circ$$ 7. **Find angle $x$ at vertex $C$:** Sum of angles in triangle is $180^\circ$: $$x = 180^\circ - 38^\circ - 31.4^\circ = 110.6^\circ$$ **Final answer:** $$x \approx 110.6^\circ$$