Question

如图所示,直线AB、CD相交于O,OE平分∠AOD,FO⊥CD于O,∠1=40°,求∠2、∠3的度数.

Answer 1

∵∠AOB=180°,∠1=40°

FO⊥CD于O,即∠COF=∠DOF=90°

∴∠3=∠AOB-∠1-∠COF=180°-40°-90°=50°

∵直线AB、CD相交于O

∴∠BOD=∠3=50°

∴∠AOD=∠AOB-∠BOD=180°-50°=130°

∵OE平分∠AOD

∴∠2=1/2∠AOD=1/2×130°=65°