Question

如图，已知角AOD：角BOD=3：4，OC平分角AOB，角COD=10度，求角AOB的度数。

Answer 1

解：设∠AOD＝3X
∵∠AOD：∠BOD＝3：4, ∠AOD＝3X
∴∠BOD＝4X
∴∠AOB＝∠AOD+∠BOD＝7X
∵OC平分∠AOB
∴∠AOC＝∠AOB/2＝7X/2＝3.5X
∴∠COD＝∠AOC-∠AOD＝0.5X
∴0.5X＝10
X＝20
∴∠AOB＝7X＝140°

Answer 2

∠AOD/∠BOD
=∠AOD/(∠BOC+∠COD)
=∠AOD/(∠AOC+∠COD)
=∠AOD/(∠AOD+∠COD+∠COD)
=∠AOD/(∠AOD+20°)=3/4
∴∠AOD=60°
∴∠BOD=80°
∴∠AOB=140°

Answer 3

∠AOB=135°