Question

已知图8中直线AB,CD,EF相交于点o，OF平分∠BOD，∠COB=∠AOC+45°，求∠AOF的度数（五分钟以内）

Answer 1

解：因为∠COB=∠AOC+45°，∠COB+∠AOC=180°，解得，∠AOC=67.5°，∠COB=112.5°，
又因为，∠BOD=∠AOC，OF平分∠BOD，∠DOF=1/2∠BOD=67.5°/2=33.75°，
又因为，∠AOD=∠BOC，
所以∠AOF=∠AOD+∠DOF=112.5°+33.75°=146.25°

Answer 2

∵∠AOC+∠COB=180°
∴∠COB=180°-∠AOC
∵,∠COB=∠AOC+45°
∴180°-∠AOC=∠AOC+45°
∴∠COB=112.5°
∵∠AOC=∠BOD，∠COB=∠AOD
∴∠BOD=67.5°，∠AOD=112.5°
∵OF平分∠BOD
∴∠DOF=33.75°
∴∠AOF=∠AOD+∠DOF=112.5°+33.75°=146.25°