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

Answer 1

解：∵C O D共线
∴∠DOE+∠COE=180°
又∵∠DOE=5∠COE
∴∠DOE=150°
又∵OE垂直AB
∴ ∠AOE=90°
∴ ∠AOD=360°-150°-90°=120°

Answer 2

答案是120°
∠COE+∠DOE=180° 因为∠DOE=5∠COE
∠COE+5∠COE=180° 所以∠COE=30° ∠DOE=5∠COE=150°
∠BOD=∠DOE-90°=60°
∠AOD=180°-∠BOD=180°-60°=120°