Question

如图,两直线AB,CD相交于点O,OE平分∠BOD,如果∠AOC:∠AOD=7:11。若OF⊥OE,求∠COF

https://gss0.baidu.com/7LsWdDW5_xN3otqbppnN2DJv/%DC%F2%DC%DF%C8%F5%CB%AE/pic/item/d606b51d3a5e9134dab4bd95.jpg

Answer 1

解：做哗∵∠AOC:∠AOD=7:11，而∠AOC+∠AOD=180°
∴∠AOC=70°，∠AOD=110°，
∴∠BOD=∠AOC=70°，又稿闭∵OE平纯敬行分∠BOD
∴∠BOE=35°，OF⊥OE，
∴∠FOE=90°，
∴∠AOF=180°-90°-35°=55°
∴∠COF=∠AOC+∠AOF=70°+55°=125°

Answer 2

∠AOC:∠AOD=7:11求出∠AOC=70°，∠AOD=110°，
∠AOC=∠携告BDO=∠BOE+∠DOE，信陪因为∠BOE=∠DOE，所以∠滑隐蠢BOE=35°，因为OF⊥OE，所以∠EOF=90°，所以∠AOF=180°-35°-90°=55°，∠COF=∠AOC+∠AOF=70+55=125°