Question

如图,AB为一条直线,OC是角AOD的平分线,OE在角BOD内,角DOE=1/3角BOD,角CDE=72度，求角EOB的度数

Answer 1

你打错了吧，应该是角COE吧

Answer 2

∠COE=∠COD+∠DOE
=1/2∠AOD+1/3∠BOD
=1/2（180°-∠BOD)+1/3∠BOD
=90°-1/6∠BOD
=72°
计算得;∠BOD=108°
∠DOE=1/3∠BOD=36°
∠EOB=∠BOD-∠DOE=72°