如图所示,在\odot O上取点D,连接AD,BD,
∵\angle AOB= 122^{{\circ} },
∴\angle ADB= \dfrac{1}{2}\angle AOB= \dfrac{1}{2}\times 122^{{\circ} }= 61^{{\circ} }.
∵四边形ADBC是圆内接四边形,
∴\angle ACB= 180^{{\circ} }-61^{{\circ} }= 119^{{\circ} }.
故答案为:119.
如图所示,在\odot O上取点D,连接AD,BD,
∵\angle AOB= 122^{{\circ} },
∴\angle ADB= \dfrac{1}{2}\angle AOB= \dfrac{1}{2}\times 122^{{\circ} }= 61^{{\circ} }.
∵四边形ADBC是圆内接四边形,
∴\angle ACB= 180^{{\circ} }-61^{{\circ} }= 119^{{\circ} }.
故答案为:119.