From the question,
[tex]\begin{gathered} m\angle AFE=m\angle BFC\text{ (Vertically opposite angles)} \\ \therefore \\ m\angle AFE=70^{\circ} \end{gathered}[/tex]We also have
[tex]m\angle AFB=m\angle EFC\text{ (Vertically opposite angl}es)[/tex]Remember that the sum of angles at a point equals 360°. Therefore
[tex]\begin{gathered} m\angle AFB+m\angle BFC+m\angle CFE+m\angle AFE=360 \\ \therefore we\text{ have} \\ 2(m\angle AFB)+2(70)=360 \\ 2(m\angle AFB)=360-140=220 \\ m\angle AFB=\frac{220}{2}=110 \end{gathered}[/tex]Therefore, m(AB) is 110°.
Hence, OPTION B is correct.
