题目内容：

∠AOB内有射线OC,OD,∠AOD=35°,∠COB=44°,且∠AOC=2/3∠DOB,求∠AOB的度数.

优质解答

∠AOC=2/3∠DOB
∠DOB-∠AOC=∠DOB-2/3∠DOB=∠DOB/3
∠AOD=∠AOC+∠COD=35° 1
∠COB=∠COD+∠DOB=44° 2
2-1得
∠DOB-∠AOC=∠DOB/3=44°-35°=9°
∠DOB=27°
∠AOC=27°-9°=18°
则∠COD=44°-27°=17°
1+2得
∠AOD+∠COB=∠AOD+2∠COD+∠DOB=∠AOB+∠COD=44°+35°=79°
∠AOB=79°-17°=62°