Question

如图,直线AB,CD交于点O,OE平分∠AOD,OF⊥OE于O，若∠BOC=80°，求∠COF的度数

如看不懂请说

Answer 1

解:因为∠BOC=80°
所以∠AOD=80°
因为OE平分∠AOD
所以∠AOE=1/2∠AOD
=40°
因为OF⊥OE于O
所以∠EOF=90°
所以∠AOF=∠EOF-∠AOE
=50°
所以∠COF=∠DOC-∠AOD-∠AOF
=50°

Answer 2

50