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+∠COE
∴∠AOD=∠EOB+∠COE=90°+30°=120°
∴∠AOD为120°

Answer 2

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