AB is a chord that passes through the center O, which means the central angles BOC and AOC are supplementary. So
[tex]m\angle BOC+m\angle AOC=180^\circ\implies m\angle AOC=180^\circ-55^\circ=125^\circ[/tex]
Central angles and the arcs they subtend have the same measure, so that
[tex]m\widehat{AC}=m\angle AOC=125^\circ[/tex]
[tex]m\widehat{CBA}=m\widehat{BC}+m\widehat{AB}=m\angle BOC+m\angle AOB=55^\circ+180^\circ=235^\circ[/tex]
