入土,OC平分∠AOB,OD是∠BOC内的一条射线

DOE=EOC+COD=20°+20°=40°2）设COA=X,则COE=X/2,BOD=40°+X/2=COD得到EOD=COD-COE=40°3）AOB=a,则得到DOE=a/2,将2）中的40°

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设∠DOE=x，∵∠DOE=13∠AOE，∴∠AOD=4∠DOE=4x，∵OD平分∠AOB，∴∠AOD=∠DOB=4x，∵OC平分∠BOD，∴∠DOC=∠COB=2x，∴∠COE=2x+x=45°，∴

∵OD平分∠AOB∠AOD=∠DOB∵OC平分∠BOD∴∠DOC=∠COB∴∠BOC+∠COD=∠BOD=2∠COD∵∠DOE=1／3∠AOE∴3∠DOE=∠AOE∴∠AOE+∠DOE=4∠EOD=∠

∵OD平分∠AOC，∴∠AOC=2∠DOC，∵∠COD=25°，∴∠AOC=50°，∵OC是∠AOB的平分线，∴∠AOB=2∠AOC，∴∠AOB=50°×2=100°．

∠EOD=45°∠EOD=∠COD+∠COE=1/2(∠AOC+∠BOC)=1/2∠AOB=1/2*90°=45°很高兴为您答题,

1、∵∠AOC＝30,∠BOC＝90∴∠AOB＝∠AOC+∠BOC＝30+90＝120∵OD平分∠AOB∴∠AOD＝∠AOB/2＝120/2＝60∵OE平分∠AOC∴∠AOE＝∠AOC/2＝30/2＝

∵OD平分∠BOC,OE平分∠AOC∴∠COD=1/2∠BOC∠COE=1/2∠AOC∴∠DOE=∠COD+∠COE=1/2(∠BOC+∠AOC)∵∠AOC+∠BOC=∠AOB=90°∴∠DOE=45

因为OD平分∠BOC,OE平分∠AOC,所以∠DOC=二分之一∠BOC∠COE=二分之一∠COA所以∠EOD=二分之一∠AOB=二分之一乘以90=45°

角DOC=1/2角AOC.角EOC=1/2角BOC.角DOE=角DOC+角EOC角AOB=角AOC+角BOC.所以角DOE=1/2角AOB给分吧

设∠BOC=x°，则∠AOB=90°+x°，∵OD平分∠AOB，∴∠BOD=12（90-x）°，∵OE平分∠BOC，∴∠BOE=12x°，∴∠DOE=∠BOD-∠BOE=12（90-x）°-12x°=

(1)∠EOF=∠EOC+∠COD+∠DOF∠AOC+∠BOD=∠AOB-∠COD=140-30=110∠EOC+∠DOF=1/2(∠AOC+∠BOD)=55∠EOF=30+55=85(2)∠EOF=

/>∵∠AOB＝90,OC平分∠AOB∴∠AOC＝∠BOC＝∠AOB/2＝90/2＝45∵OD平分∠BOC∴∠COD＝∠BOD＝∠BOC/2＝45/2＝22.5∵OE平分∠AOC∴∠COE＝∠AOE＝

1、OD靠近OC∵OC平分∠AOB∴∠BOC＝∠AOB/2∵OD是∠BOC内的三等分线∴∠COD＝∠BOC/3∴∠COD＝1/3×∠AOB/2＝∠AOB/6∴∠AOB/∠COD＝6∴∠AOB是∠COD

（1）因为OC平分∠AOB,所以∠AOC=∠BOC=1/2∠AOB=80°因为∠AOD=70°,所以角COD=∠AOC-∠AOD=10°（2）因为OC平分∠AOB,所以∠AOC=∠BOC=1/2∠AO

∵∠COD=28°,OC,OD是∠AOB的三等分线∴∠AOB=3∠COD=84°又∵OE平分∠AOB∴∠BOE=42°

因为OC平分<AOB 所以<AOC=<BOC=1/2<AOB因为OD平分< AOC 所以<AOD=<COD=1/2<AOC

1.∠EOF=∠EOC+∠COD+∠DOF=0.5∠AOC+∠COD+0.5∠DOB=0.5(∠AOC+∠DOB)+∠COD=0.5(∠AOB-∠COD)+∠COD=0.5*(120°-20°）+20

设∠BOE=x,∠COD=y则∠COE=x,∠BOD=2x+y,∠AOD=y∵OD平分∠AOB∴90-y=2x+y∴2x+2x=90∴x+y=45°∴∠DOE=45°请采纳回答

设∠BOE=x,∠COD=y则∠COE=x,∠BOD=2x+y,∠AOD=y∵OD平分∠AOB∴90-y=2x+y∴2x+2x=90∴x+y=45°∴∠DOE=45°