如图,coe=90,aof=30,of平分aoc,求boe的度数
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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∴