Question

For circle O, mCD=125° and m

Answer 1

CA is a diameter of the circle, so $m\widehat{AC}=180^\circ$, which means $m\angle AOB=m\angle AOD=m\widehat{AD}=180^\circ-m\widehat{CD}=55^\circ$. Then $m\boxed{\angle ABO}=90^\circ-55^\circ=35^\circ$.

This means $m\angle CBO=55^\circ-35^\circ=20^\circ$. Also, if $m\angle AOB=55^\circ$, then $m\angle BOC=180^\circ-55^\circ=125^\circ$, which in turns tells us that $m\boxed{\angle BCO}=180^\circ-20^\circ-125^\circ=35^\circ$.