如图,已知直线AB,CD,EF相交于O点,OF平分∠BOD,∠COB=∠AOC+45°,求∠AOF的度数?

∵∠AOC+∠COB=180°∴∠COB=180°-∠AOC∵,∠COB=∠AOC+45°∴180°-∠AOC=∠AOC+45°∴∠AOC=67。5°=∠BOD∴∠COB=112。5°=∠AOD∵OF平分∠BOD∴∠DOF=33。75°∴∠AOF=∠AOD+∠DOF=112。5°+33。75°=146。25°