如图,射线OB和射线OE分别是角AOC和角DOF的角平分线,已知角AOF等于135度,角BOE比角COD的2倍大12度,求角BOE和角COD的大小
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如图,射线OB和射线OE分别是角AOC和角DOF的角平分线,已知角AOF等于135度,角BOE比角COD的2倍大12度,求角BOE和角COD的大小
如图,射线OB和射线OE分别是角AOC和角DOF的角平分线,已知角AOF等于135度,角BOE比角COD的2倍大12度,
求角BOE和角COD的大小
如图,射线OB和射线OE分别是角AOC和角DOF的角平分线,已知角AOF等于135度,角BOE比角COD的2倍大12度,求角BOE和角COD的大小
∠BOE
=∠BOC+∠COD+∠DOE
=1/2∠AOC+∠COD+1/2∠DOF
=1/2(∠AOC+∠COD+∠DOF)+1/2∠COD
=1/2∠AOF+1/2∠COD
=135°/2+1/2∠COD
依题意得:∠BOE=2∠COD+12°
∴135°/2+1/2∠COD=2∠COD+12°
解得:∠COD=37°
从而∠BOE=37°×2+12°=86°
因为OB、OE分别是∠AOC和∠DOF的角平分线
所以∠AOC=2∠BOC,∠DOF=2∠DOE
又因为∠AOF=∠AOC+∠COD+∠DOF
=2(∠BOC+∠DOE)+∠COD=135.................(1)
由题可知,∠BOD=∠BOC+∠DOE+∠COD=2∠COD+12....(2)
联立(1...
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因为OB、OE分别是∠AOC和∠DOF的角平分线
所以∠AOC=2∠BOC,∠DOF=2∠DOE
又因为∠AOF=∠AOC+∠COD+∠DOF
=2(∠BOC+∠DOE)+∠COD=135.................(1)
由题可知,∠BOD=∠BOC+∠DOE+∠COD=2∠COD+12....(2)
联立(1)(2)解得∠COD=37
∠BOE=86
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∠BOE
=∠BOC+∠COD+∠DOE
=1/2∠AOC+∠COD+1/2∠DOF
=1/2(∠AOC+∠COD+∠DOF)+1/2∠COD
=1/2∠AOF+1/2∠COD
=135°/2+1/2∠COD
依题意得:∠BOE=2∠COD+12°
∴135°/2+1/2∠COD=2∠COD+12°
解得:∠COD=37°
从...
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∠BOE
=∠BOC+∠COD+∠DOE
=1/2∠AOC+∠COD+1/2∠DOF
=1/2(∠AOC+∠COD+∠DOF)+1/2∠COD
=1/2∠AOF+1/2∠COD
=135°/2+1/2∠COD
依题意得:∠BOE=2∠COD+12°
∴135°/2+1/2∠COD=2∠COD+12°
解得:∠COD=37°
从而∠BOE=37°×2+12°=86
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因为OB、OE分别是∠AOC和∠DOF的角平分线
所以∠AOC=2∠BOC,∠DOF=2∠DOE
又因为∠AOF=∠AOC+∠COD+∠DOF
=2(∠BOC+∠DOE)+∠COD=135.................(1)
由题可知,∠BOD=∠BOC+∠DOE+∠COD=2∠COD+12....(2)
联立(1...
全部展开
因为OB、OE分别是∠AOC和∠DOF的角平分线
所以∠AOC=2∠BOC,∠DOF=2∠DOE
又因为∠AOF=∠AOC+∠COD+∠DOF
=2(∠BOC+∠DOE)+∠COD=135.................(1)
由题可知,∠BOD=∠BOC+∠DOE+∠COD=2∠COD+12....(2)
联立(1)(2)解得∠COD=37
∠BOE=86
收起