如图2.点D,E分别在BC,AC上,AD,BE相交于点F,若∠1=46°,∠2=42°,∠C=37°,求∠AFB的度数.

由题可知 ∠1+∠2+∠3+∠ABF+∠BAF=180°
∠ABF+∠BAF=180-46-42-37=55°
且∠ABF+∠BAF+∠AFB=180°
∠AFB=125°