已知:∠COB=2∠AOC,OD平分∠AOB,且∠COD=19°,求∠AOB的度数

设∠AOC=X，则∠COB=2X，
∠AOB=∠COB+∠AOC=3X.
由题可知，∠COD=∠AOD-∠AOC=19°
且 OD平分∠AOB
∴∠AOD=1/2∠AOB=3/2X
∠AOD-∠AOC=19° 即 3/2X-X=19°
解得 X=38°
∠AOB=3X=38°*3=114°

解设角AOC=x
则∠BOC=2x
则∠AOB=3x
则∠AOD=3x/2
则∠COD=∠AOD-∠AOC=3x/2-x=x/2
又由∠COD=19°
知x=38°
则∠AOB=3x=114°

设∠AOC=x,则∠BOC=2x．
∴∠AOB=∠AOC∠BOC=3x
又OD平分∠AOB,
∴∠AOD=1.5x．
∴∠COD=∠AOD-∠AOC=1.5x-x=19°
∴x=38°
∴∠AOB=114°．
故答案为114°．

∠AOD=1/2∠AOB
∠AOC+∠COB=∠AOB
3∠AOC=∠AOB
∠AOC=1/3∠AOB
∠AOD=1/2∠AOB
∠COD=1/6∠AOB=19º
∠AOB=114º

∵∠AOD-∠AOC=∠COD=19º
∠AOC=1/3∠AOB
∠AOD=1/2∠AOB
∴1/2∠AOB-1/3∠AOB=19º
∴1/6∠AOB=19º
∴∠AOB=114º