如图,已知直线AB,OC⊥AB,OD⊥OE,若∠COE=1/5∠BOD,则求∠COE,∠BOD,∠AOE的度数.
问题描述:
如图,已知直线AB,OC⊥AB,OD⊥OE,若∠COE=
∠BOD,则求∠COE,∠BOD,∠AOE的度数.1 5
答
∵OC⊥AB,OD⊥OE,
∴∠DOE=∠AOC=90°,
∵∠COE+∠DOC=∠DOE=90°,
∠AOD+∠DOC=∠AOC=90°,
∴∠COE=∠AOD,
∵∠BOD=180°-∠AOD,
∵∠COE=
∠BOD,1 5
∴∠COE=30°,
∴∠BOD=180°-∠AOD
=180°-∠COE
=180°-30°
=150°;
∴∠AOE=∠AOC+∠COE
=90°+30°
=120°.