Answer:
145°
Step-by-step explanation:
As per the given information: quadrilateral ABCD is a parallelogram.
[tex] m\angle B + m\angle DCB =180\degree \\[/tex]
(adjacent angles of a parallelogram)
[tex] 110\degree + m\angle DCB =180\degree \\
m\angle DCB =180\degree - 110\degree \\
m\angle DCB =70\degree\\
m\angle DCG=\frac{1}{2} m\angle DCB \\(\because GC \: bisects \: \angle DCB) \\
m\angle DCF=\frac{1}{2}\times 70\degree... (\because C-G-F) \\
m\angle DCF=35\degree\\
m\angle BFC = m\angle DCF = 35\degree\\ (Alternate \: \angle s) \\\\
m\angle BFC + m\angle CFA = 180\degree\\ (straight \: line \: \angle s) \\\\
35\degree + m\angle CFA= 180\degree\\
m\angle CFA = 180\degree-35\degree \\\\
m\angle CFA= 145\degree \\\\
\huge\purple {\therefore m\angle GFA = 145\degree} \\(\because C-G-F) \\[/tex]
