Question

如图所示,角COB=2角AOC,OD平分角AOB，角COD=19°，求角AOB的度数

Answer 1

∵OD平分∠AOB

∴∠BOD=∠AOD

∵∠COB=2∠AOC

∴∠BOD+∠COD=2∠AOC

∵∠AOC=∠AOD-∠COD

又∠COD=19°

∴∠BOD+∠COD=2(∠AOD-∠COD)

∴∠BOD=∠AOD=3∠COD=57°

∴∠AOB= 2∠BOD=114°

Answer 2

∵OD平分∠AOB
∴∠BOD=∠AOD
∵∠COB=2∠AOC
∴∠BOD+∠COD=2∠AOC
∵∠AOC=∠AOD-∠COD
又∠COD=19°
∴∠BOD+∠COD=2(∠AOD-∠COD)
∴∠BOD=∠AOD=3∠COD=57°
∴∠AOB= 2∠BOD=114°

Answer 3

∵OD平分∠AOB
∴∠BOD=∠AOD
∵∠COB=2∠AOC
∴∠BOD+∠COD=2∠AOC
∵∠AOC=∠AOD-∠COD
又∵∠COD=19°
∴∠BOD+∠COD=2(∠AOD-∠COD)
∴∠BOD=∠AOD=3∠COD=57°
∴∠AOB= 2∠BOD=114°

Answer 4

404

Answer 5

114