Question

如图所示，已知∠COB=2∠AOC,OD平分∠AOB，且∠COD=19°的度数，求∠AOB的度数，要有解释

Answer 1

因为角COD=20度20分，OD平分角AOB ,，所以角AOB=2角AOD=2(角COD+角AOC),因为角BOC=2角AOC，角AOB=角BOC+角AOC，所以角AOB=3角AOC,即角AOC=角AOB/3，所以角AOB=2(20度20分+角AOB/3)，所以角AOB=122度简单∠BOC+∠AOC=∠AOB=3∠AOC,OD平分∠AOB，所以∠AOD=3/2∠AOC ∠AOD-∠AOC=1/2∠AOC=20°所以∠AOC=40 得到∠AOB=120° 设∠AOC=α， 赞同
设∠AOC=a，则∠AOB=3a，∠COD=(3/2)a-a=(1/2)a=20°20′所以∠AOB=3a=122°
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Answer 2

∠AOC = 1/3∠AOB
∠AOD = 1/2∠AOB
∴∠COD =∠AOD-∠AOC = 1/6∠AOB
∠AOB = 19÷（1/2-1/3）=114

Answer 3

∠AOC = 1/3∠AOB
∠AOD = 1/2∠AOB
∴∠COD =∠AOD-∠AOC = 1/6∠AOB
∠AOB = 19÷（1/2-1/3）=114

追问

厉害，没图都能写出来

Answer 4

解：∵OD平分∠AOB
∴∠BOD=∠DOA=1/2∠AOB
∵∠COD=∠AOD-∠AOC = 1/6∠AOB
∴∠AOB=6∠COD=6×19°=144°