1. **Problem Statement:** We are given triangle MNP with incenter C. We know the length of side MP is 11, angle M is 20°, angle N is 30°, and segment CN (from vertex N to incenter C) is 4. We need to find the length MC. 2. **Relevant Theorem:** The incenter is the intersection of the angle bisectors of a triangle. The incenter is equidistant from all sides of the triangle, and the segments from vertices to the incenter relate to the sides and angles. 3. **Step 1: Find angle P.** Since the sum of angles in a triangle is 180°, $$\angle P = 180^\circ - 20^\circ - 30^\circ = 130^\circ.$$ 4. **Step 2: Use the Law of Sines to find sides MN and NP.** Let side lengths be: - $MP = 11$ (given) - $MN = a$ - $NP = b$ By Law of Sines: $$\frac{MP}{\sin \angle N} = \frac{MN}{\sin \angle P} = \frac{NP}{\sin \angle M}.$$ Calculate the common ratio $k$: $$k = \frac{MP}{\sin 30^\circ} = \frac{11}{0.5} = 22.$$ Then, $$MN = k \sin 130^\circ = 22 \times \sin 130^\circ = 22 \times 0.7660 = 16.852,$$ $$NP = k \sin 20^\circ = 22 \times 0.3420 = 7.524.$$ 5. **Step 3: Use the formula for the distance from vertex to incenter:** The distance from vertex $N$ to incenter $C$ is given by $$NC = \frac{r}{\sin(\frac{\angle N}{2})},$$ where $r$ is the inradius (distance from incenter to sides). Given $NC = 4$, and $\angle N = 30^\circ$, so $$4 = \frac{r}{\sin 15^\circ} \implies r = 4 \times \sin 15^\circ = 4 \times 0.2588 = 1.035.$$ 6. **Step 4: Find $MC$ using the same formula for vertex M:** $$MC = \frac{r}{\sin(\frac{\angle M}{2})} = \frac{1.035}{\sin 10^\circ} = \frac{1.035}{0.1736} = 5.96.$$ 7. **Final answer:** \boxed{MC \approx 5.96}