Question

如以下图：已知∠AOB=90°,∠AOC=60°,OD平分∠BOC,OE平分∠AOC.求∠DOE的度数

Answer 1

∠AOB=90°,
∠AOC=60°,
∠BOC=∠AOB+∠AOC=90°+60°=150°
OD平分∠BOC,
∠COD=∠BOC/2 = 150°/2 = 75°,
OE平分∠AOC.
∠COE=∠AOC/2 = 60°/2 = 30°,
∠DOE = ∠COD-∠COE = 75°+30° = 45°

Answer 2

∠BOD=(90+60)/2=75度

则∠AOD=90-75=15度
∠DOE=∠AOD+∠AOE=15+60/2=45度

Answer 3

因为OD平分∠BOC，所以∠DOC=∠BOD=∠BOC/2
∠BOC=∠BOA+∠AOC=90+60=150

∠DOC=75

∠DOA=∠DOC-∠AOC=15

∠AOE=∠AOC/2=30

所以∠DOE=∠DOA+∠AOE=45

Answer 4

∵∠BOC=∠AOB+∠AOC=90°+60°=150°
OD平分∠BOC,OE平分∠AOC
∴COD=∠BOD=1/2∠BOC=1/2×150°=75°
∠COE=∠AOE=1/2∠AOC=1/2×60°=30°
∴∠DOE=∠COD-∠COE=75°-30°=45°