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
网友回答
∵∠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°
======以下答案可供参考======
供参考答案1:
∠AOD=∠BOC
=∠BOE+∠COE
=90+1/6∠COD
=90+1/6*180
=120度