In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $\displaystyle \angle \mathrm{CBA}$ and $\displaystyle \angle \mathrm{BCO}$, if $\displaystyle \angle \mathrm{PCA}=110^{\circ}$.

\[\begin{aligned} & \angle \mathrm{ACB}=\angle \mathrm{OCB}+\angle \mathrm{OCA}=90^{\circ} \\ & \angle \mathrm{PCB}+\angle \mathrm{OCB}+\angle \mathrm{OCA}=110^{\circ} \\ & \angle \mathrm{PCB}=110^{\circ}-90^{\circ}=20^{\circ} \\ & \angle \mathrm{PCB}+\angle \mathrm{OCB}=90^{\circ} \\ & \angle \mathrm{OCB}=90^{\circ}-20^{\circ}=70^{\circ} \\ & \mathrm{As} \mathrm{OB}=\mathrm{OC} \Rightarrow \angle \mathrm{OBC}=\angle \mathrm{OCB} \\ & \angle \mathrm{OBC}=\angle \mathrm{OCB}=70^{\circ} \end{aligned} \]