Ответ:

∠CAO+∠AOC+∠OCA=180° <=>

∠CAB/2 +∠AOC +∠BCA/2 =180° <=>

∠CAB+∠BCA = 2(180°-∠AOC)

∠CAB+∠ABC+∠BCA=180° <=>

∠ABC= 180°-(∠CAB+∠BCA) <=>

∠ABC= 180°-2(180°-∠AOC) = 2(∠AOC-90°)

∠ABC= 2(124°-90°) =68

∠AOC=124°