Question

如图，角BOC=2AOC，OD平分角AOB，角COD=20°，求角AOC的度数

大师

Answer 1

∵OD平分角AOB
∴角AOD=1/2角AOB=1/2（角BOC+角AOC）=角COD+角AOC
∵角BOC=2AOC
∴1/2（2角AOC+角AOC）=角COD+角AOC
角AOC+1/2角AOC=20°+角AOC
角AOC=40°

Answer 2

解：
∵∠BOC＝2∠AOC
∴∠AOB＝∠AOC+∠BOC＝∠AOC+2∠AOC＝3∠AOC
∵OD平分∠AOB
∴∠AOD＝∠AOB/2＝3∠AOC/2
∴∠COD＝∠AOD-∠AOC＝3∠AOC/2-∠AOC＝∠AOC/2
∵∠COD＝20°
∴∠AOC/2＝20°
∴∠AOC＝40°
∴∠AOB＝3∠AOC＝120°