Question

数学题 帮帮忙 O(∩_∩)O谢谢 急啊 跪求大家 写算式

如图13-2，OD平分∠AOC，∠AOB=3∠COD，∠BOC=4∠AOD，求∠AOB的度数

Answer 1

∵OD平分∠AOC
∴∠AOD=∠COD=½∠AOC
∵∠AOB=3∠COD，∠BOC=4∠AOD
∴∠AOB=3∠COD,∠BOC=4∠COD
∵∠AOB+∠BOC+∠AOC=360°
∴3∠COD+4∠COD+2∠COD=360°
即∠COD=40°
∴∠AOB=3×40°=120°

Answer 2

∵OD平分∠AOC
∴∠COD＝∠AOD
∵∠AOB=3∠COD，∠BOC=4∠AOD,
∴∠AOB＝∠BOC＝∠AOC＝（2+3+4）∠AOD＝360°
∴∠AOD＝40°
∴∠AOB＝120°

Answer 3

OD平分∠AOC
说明 ∠COD=∠AOD 可以令它们都为a
∠AOB=3∠COD=3a，∠BOC=4∠AOD=4a
∠COD+∠AOD +∠AOB+∠BOC=180 即a+a+3a+4a=360 解得a=40
∠AOB=3x40=120度

Answer 4

9∠AOD=360°
∠AOD=40°
∠AOB=120°

Answer 5

AOB+BOC+COD+AOD=360°

COD=40

AOB=120

Answer 6

解：设∠COD=x，
OD平分∠AOC ∠AOD=∠COD=x
∠AOB=3∠COD=3x
∠BOC=4∠AOD=4x
由图可知∠AOD+∠COD+∠AOB+∠BOC=360°=9x
解得x=40°
∠AOB=3x=120°