Question

如图所示，直线AB，CD相交于点O，OE垂直AB于O，且∠DOE=5∠COE，求∠AOD的度数

如图

Answer 1

解：∵∠DOE=5∠COE，∠DOE+∠COE=180°
∴∠COE=30°
∵OE垂直AB于O
∴∠EOB=90°
∵∠BOC=∠AOD且∠BOC=∠EOB+∠辩如COE
∴∠雀灶好AOD=∠EOB+∠COE=90°+30°=120°
∴∠顷铅AOD为120°

Answer 2

∠卖橘AOD=∠BOC
=∠BOE+∠COE
=90+1/中仔团6∠戚皮COD
=90+1/6*180
=120度