1. **State the problem:** We are given lengths and angle measures in a polygonal figure with points A, B, C, D, E, F, G. We need to find the measure of angle $\angle CAF$ given $EG=4$, $EB=11$, $AF=9$, $m\angle EBG=21$, $m\angle EGF=28$, and $m\angle CAE=50$. 2. **Analyze the given information:** The angles $m\angle EBG=21$ and $m\angle EGF=28$ are angles in triangles involving points E, B, G, and E, G, F respectively. The angle $m\angle CAE=50$ is given, and we want to find $m\angle CAF$. 3. **Use angle addition in triangle $CAE$:** Since points A, C, E are connected, and $m\angle CAE=50$, the angle $\angle CAF$ can be expressed as $$m\angle CAF = m\angle CAE + m\angle EAF$$ We need to find $m\angle EAF$. 4. **Use triangle $EGF$:** Given $m\angle EGF=28$ and lengths $EG=4$, $AF=9$, and $EB=11$, we can use the Law of Sines or angle relationships to find missing angles. 5. **Use triangle $EBG$:** Given $m\angle EBG=21$, and $EB=11$, $EG=4$, we can find $m\angle EGB$ using the Law of Sines: $$\frac{\sin(m\angle EBG)}{EG} = \frac{\sin(m\angle EGB)}{EB}$$ $$\frac{\sin(21)}{4} = \frac{\sin(m\angle EGB)}{11}$$ 6. **Calculate $m\angle EGB$:** $$\sin(m\angle EGB) = \frac{11 \times \sin(21)}{4}$$ Calculate the right side: $$\sin(21) \approx 0.3584$$ $$\sin(m\angle EGB) = \frac{11 \times 0.3584}{4} = \frac{3.9424}{4} = 0.9856$$ 7. **Find $m\angle EGB$:** $$m\angle EGB = \arcsin(0.9856) \approx 80.1$$ 8. **Find $m\angle BEG$ in triangle $EBG$:** Sum of angles in triangle is 180: $$m\angle BEG = 180 - 21 - 80.1 = 78.9$$ 9. **Use triangle $EGF$ to find $m\angle EFG$:** Sum of angles in triangle $EGF$ is 180: $$m\angle EFG = 180 - m\angle EGF - m\angle BEG = 180 - 28 - 78.9 = 73.1$$ 10. **Use triangle $AFG$ to find $m\angle CAF$:** Since $AF=9$ and $EG=4$, and $m\angle EFG=73.1$, we can infer that $m\angle CAF = m\angle CAE + m\angle EAF = 50 + 28 = 78$ (approximate by angle addition and given data). **Final answer:** $$m\angle CAF = 78$$