解:作∠BAC的角平分线与CO的延长线交于点D,连接BD, ∵∠BAD=∠DAC,AB=AC,AD=AD, ∴△ABD≌△ACD, ∴BD=CD,∠ABD=∠ACD, ∴∠DBC=∠DCB, ∵∠BAC=80°(已知), ∴∠ABC=∠ACB=50°(三角形内角和定理); 又∠OCA=20°, ∴∠ABD=∠ACD=20°, ∠OBD=∠ABC﹣∠ABD﹣∠OBC=50°﹣20°﹣10°=20°=∠ABD, ∠DOB=∠OBC+∠OCB=∠OBC+∠ABC﹣∠ACO=10°+50°﹣20°=40°=∠BAD, ∴∠OBD=∠ABD,∠DOB=∠DAB,BD=BD, ∴△ABD≌△OBD, ∴AB=OB, ∴∠BAO=∠AOB, ∴∠BAO=(180°﹣∠ABO) =[180°﹣(∠ABC﹣∠OBC)] =(180°﹣40°)=70°. |