入土,OC平分∠AOB,OD是∠BOC内的一条射线
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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°