Answer:
55°
Step-by-step explanation:
Let the centre of the circle be C
mFG=110 (It's hard to read but I assume thats what the 110 marks)
mFG=<FCG=110
Consider ΔFCG (an isosceles triangle since CF=CG=radius). Angles in a triangle add to 180 and <CFG=<CGF so <CFG=(1/2)(180-110)=(1/2)(70)=35°
<EFG+<CFG=90
<EFG+35=90
<EFG=55°
