1. **State the problem:** We have a regular pentagon ABCDE and a point F outside the pentagon such that the angle \(\angle AEF = 96^\circ\). We need to find the size of the obtuse angle \(\angle FED\). 2. **Recall properties of a regular pentagon:** Each interior angle of a regular pentagon is \(108^\circ\) because the sum of interior angles in any pentagon is \( (5-2) \times 180 = 540^\circ\), and \(540^\circ / 5 = 108^\circ\). 3. **Analyze angles around point E:** At vertex E, we know \(\angle AEF = 96^\circ\). The points A, E, and D are vertices of the pentagon, so \(\angle AED = 108^\circ\) (interior angle). 4. **Understand the layout:** Since F is outside the pentagon extending vertically from E, angles around E formed by lines EA, EF, and ED sum to \(360^\circ\) because they meet at a point. 5. **Calculate the angle \(\angle FED\):** We have the three angles around point E on that side: \(\angle AEF = 96^\circ\), \(\angle AED = 108^\circ\), and \(\angle FED\) which is what we want. Since these three angles are adjacent around point E on a straight path, the sum is \(360^\circ\): $$\angle AEF + \angle FED + \angle AED = 360^\circ$$ Plugging in known values: $$96^\circ + \angle FED + 108^\circ = 360^\circ$$ Simplify: $$\angle FED = 360^\circ - 96^\circ - 108^\circ = 156^\circ$$ 6. **Interpretation:** The angle \(\angle FED = 156^\circ\) is obtuse as requested. **Final answer:** \(\boxed{156^\circ}\)