如图,已知直线AB、CD、EF相交于点O,OF平分∠BOD,∠COB=∠AOC+40°,求∠AOF

2025-06-26 21:46:22
推荐回答(2个)
回答1:

∵∠AOC+∠COB=180°
∴∠COB=180°-∠AOC
∵,∠COB=∠AOC+40°
∴180°-∠AOC=∠AOC+40°
∴∠AOC=70°=∠BOD
∴∠COB=110°=∠AOD
∵OF平分∠BOD
∴∠DOF=35°
∴∠AOF=∠AOD+∠DOF=110°+35°=145°

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回答2:

由图:∠COB+∠AOC=180°,又因:∠COB=∠AOC+45° 代入
∠AOC+45°+∠AOC=180°,
解:∠AOC=67.5°  ,∠COB=∠AOC+45°=112.5°
因OF平分∠BOD,即OE平分∠AOC,∠EOC=∠EOA=1/2∠AOC=33.75°
又因,∠EOC与,∠FOD对顶角所以,∠FOD=33.75° 
∠COB与∠AOD对顶角,所以∠AOD=∠COB=112.5°
∠AOF=∠AOD+∠FOD=112.5°+33.75° =146.25°