1. **Problem statement:** Given that $\angle FEB$ and $\angle CFD$ are straight angles (each 180°), and $\angle EFC$ is a right angle (90°), find the measure of $\angle CFB$.\n\n2. **Given information:**\n- $x = 11$\n- $\angle FE A = 3x^\circ = 3 \times 11 = 33^\circ$\n- $\angle AFD = (4x + 13)^\circ = 4 \times 11 + 13 = 44 + 13 = 57^\circ$\n- $\angle EFC = 90^\circ$ (right angle)\n\n3. **Find $\angle CFB$:**\n\n4. Since $\angle FEB$ is a straight angle, $\angle FEB = 180^\circ$. The ray $FE$ is vertical up, $FC$ is horizontal left, so $\angle EFC = 90^\circ$ is the right angle between $FE$ and $FC$.\n\n5. The diagonal line $FB$ passes through $F$ and forms $\angle CFB$ with $FC$. We want to find $m\angle CFB$.\n\n6. Note that $\angle FEB = 180^\circ$ is composed of $\angle EFA + \angle AFB + \angle BFD$ but we focus on the angles around $F$.\n\n7. The angles around point $F$ sum to 360°: $\angle EFC + \angle CFB + \angle BFD + \angle DFE = 360^\circ$.\n\n8. From the problem, $\angle FE A = 33^\circ$ and $\angle AFD = 57^\circ$. The diagonal $FB$ splits the straight angle $FEB$ into two parts: $\angle EFB = 3x = 33^\circ$ and $\angle BFD = 4x + 13 = 57^\circ$.\n\n9. Since $\angle EFC = 90^\circ$ and $\angle CFB$ is adjacent to it, and $\angle EFB = 33^\circ$, then $\angle CFB = 90^\circ - 33^\circ = 57^\circ$.\n\n10. **Final answer:**\n$$m\angle CFB = 57^\circ$$