如图,coe=90,aof=30,of平分aoc,求boe的度数

设角1为X6X=180,所以X=30.∠BOD=∠AOC=60因此,∠COE=∠COD-∠1=150.故,∠COF=75,则∠AOF=∠COF+∠AOC=135

互余,角AOC角BOD互补角AOD角BOC

∵直线EF∴∠EOF＝180∴∠EOA+∠AOF＝180∵∠AOF＝4∠EOA∴∠EOA+4∠EOA＝180∴∠EOA＝36∵∠AOC＝90∴∠EOC＝∠AOC-∠EOA＝90-36＝54

因为直线AB,CD相交与点O,OE平分B∠OD,OF平分∠COE,∠2=4∠1,∠BOD=2∠1,∠AOB=6∠1所以∠1=180÷6=30度.∠2=4∠1=120度根据对顶角相等的性质∠2=∠BOC

/>∵∠AOD：∠BOE=4:1∴设∠AOD=4x,则∠BOE=x∵OE平分∠BOD,∴∠DOE=∠BOE=x,∠BOD=2x∠AOD+∠BOD=180°4x+2x=180°x=30°∴∠BOD=2x

OE平分∠BOD∠∠BOD=2∠BOE∠AOD:∠BOE=4:1∠AOD:∠BOD=4:2=2:1∠AOD+∠BOD=180°∠AOD=120°∠BOD=60°∠BOE=∠DOE=30°∠AOC=∠B

因为,∠COE=3∠EOD,∠COD=180°∴∠EOD=45°∵∠EOB=90°∴∠BOD=90°-45°=45°

1）∠AOF,∴OF⊥AB,∴∠BOF=90°2）∠AOF=90°,∠DOF=65°,∴∠BOD=180-90-65=25°AB、CD相交,∴∠AOC=∠BOD=25°∠AOE=∠AOC+∠COE=2

∵∠AOB=90°，OC平分∠AOB∴∠BOC=12∠AOB=45°（3分）∵∠BOD=∠COD-∠BOC=90°-45°=45°∠BOD=3∠DOE（6分）∴∠DOE=15°（8分）∴∠COE=∠C

角AOF=135∠AOD:∠BOE=4:1,OE是角平分线,则AOD:BOD=2:1即角AOD=120,AOC=60;BOC=120,BOE=30；则角COE=BOC+BOE=120+30=150,O

45°因为AOC=90°所以AOD=90°因为AOF=3FOBFOB=AOE所以AOE+AOF=4AOE=180°AOE=45°所以EOC=AOC-AOE=90°-45°=45°

OE评分角BOD,∠AOD︰∠BOE=7︰1∠AOD︰∠BOD=7︰2∠AOD+∠BOD=180º∠AOD=180º×7/(7+2)=140º,∠BOD=60º

/>因为：OE⊥OF；所以：∠EOF=90°；因为：AOB是直线；所以：∠AOF+∠EOB=180°-90°=90°；因为：∠AOF=48°；所以：∠EOB=90°-∠AOF=90°-48°=42°；

∵OE平分∠BOD,∴∠DOE=∠BOE.∵∠AOD:∠BOE=7:1,∴∠AOD=7∠BOE.又∵∠AOD+∠DOE+∠BOE=180°,∴7∠BOE+∠BOE+∠BOE=180°,∴∠DOE=∠B

∵∠DOE=3∠COE又∵∠DOE+∠COE=180°∴∠COE=45°又∵∠BOE=90°∴根据图像,∠BOC=45°∴∠AOD=∠BOC=45°

1:∠aoe=∠cof=90°∠boe=∠dof=90°∠doe+∠bod=∠bof+∠bod=90°∠doe=∠bof∠aof+∠bof=180∠aof+∠doe+1802:同理:∠coe+∠doe

∵∠AOD=180°-∠AOC=180°-120°=60°,∴∠AOF=1/3∠AOD=20°,∴∠BOE=∠AOF=20°.

 再问：第三步角AOF=135度是怎么求的再答： 再问：谢了再答：不客气

121再问：过程啊！！！！%>_

/>∵直线AB∴∠AOB＝180∴∠AOF+∠BOF＝180∵∠AOF＝3∠BOF∴3∠BOF+∠BOF＝180∴∠BOF＝45∵∠BOD与∠AOC为对顶角,∠AOC＝90∴∠BOD＝∠AOC＝90∴