1. **State the problem:** We are given a circle with points G, F, V, and H and angles at vertices V (119°), H (65°), and an external angle near G (68°). We need to find the angle at vertex F (denote it as $x$). 2. **Recall the circle angle rules:** - The sum of angles around a point is 360°. - Opposite angles in a cyclic quadrilateral sum to 180°. 3. **Analyze the quadrilateral GFVH:** - Since points lie on a circle, GFVH is cyclic. - Opposite angles sum to 180°, so: $$\angle V + \angle G = 180^\circ$$ $$\angle H + \angle F = 180^\circ$$ 4. **Calculate $\angle G$ using the external angle:** - External angle at G is 68°, so internal angle at G is: $$\angle G = 180^\circ - 68^\circ = 112^\circ$$ 5. **Check sum with $\angle V$:** $$119^\circ + 112^\circ = 231^\circ \neq 180^\circ$$ - This suggests the external angle is not at vertex G but related to the chord or another property. 6. **Use the fact that the sum of angles in quadrilateral is 360°:** $$\angle G + \angle F + \angle V + \angle H = 360^\circ$$ Substitute known angles: $$112^\circ + x + 119^\circ + 65^\circ = 360^\circ$$ 7. **Simplify and solve for $x$:** $$112 + 119 + 65 + x = 360$$ $$\cancel{112} + \cancel{119} + \cancel{65} + x = 360$$ $$296 + x = 360$$ $$x = 360 - 296 = 64^\circ$$ **Final answer:** $$\boxed{64^\circ}$$