CA is a diameter of the circle, so [tex]m\widehat{AC}=180^\circ[/tex], which means [tex]m\angle AOB=m\angle AOD=m\widehat{AD}=180^\circ-m\widehat{CD}=55^\circ[/tex]. Then [tex]m\boxed{\angle ABO}=90^\circ-55^\circ=35^\circ[/tex].
This means [tex]m\angle CBO=55^\circ-35^\circ=20^\circ[/tex]. Also, if [tex]m\angle AOB=55^\circ[/tex], then [tex]m\angle BOC=180^\circ-55^\circ=125^\circ[/tex], which in turns tells us that [tex]m\boxed{\angle BCO}=180^\circ-20^\circ-125^\circ=35^\circ[/tex].
