Question

一道数学题，谢谢大家

AOC为一条直线，OD是∠AOB的平分线，OE在∠BOC内，∠BOE=1/3∠BOC，∠DOE=72°，求∠EOC的度数

Answer 1

∠AOB+∠BOC=180°
1/2∠AOB+1/3∠BOC=72°
∠AOB=72°
∠BOC=108°
∠EOC=2/3∠BOC=72°

Answer 2

设∠AOD=x,∠BOE=y.
∠DOB=∠AOD=x,∠EOC=2∠BOE=2y
x+y=∠DOB+∠BOE=∠DOE=72
x+x+y+2y=∠AOD+∠DOB+∠BOE+∠EOC=180

x=36 ,y=36
∠EOC=2y=72

Answer 3

72°
设∠BOE=X，则∠EOC=2X，∠DOB=72-X
2(72-X)+3X=180得X=36
2X=72

Answer 4

72°

Answer 5

设角AOB为x度
BOD=x/2
BOC=180-X
BOE=(180-X)/3=60-X/3
DOE=X/2+60-X/3=72
解方程
X/6=12
X=72
BOC=180-72=108
EOC=108/3*2=72
题目中省略了角符号和度数符号