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In acute triangle $ABC,\ \angle A = 68^\circ$.

Let be the circumcenter of triangle $ABC$.
Find, $\angle OBC$ in degrees.

$\text{Let $\angle OBC = {\color{red}x}$ }$

$\begin{array}{|rcll|} \hline 2{\color{red}x} + 2{\color{blue}y} + 2{\color{green}z} &=& 180^{\circ} \quad | \quad : 2 \\ {\color{red}x} + {\color{blue}y} + {\color{green}z} &=& 90^{\circ} \quad | \quad {\color{blue}y} + {\color{green}z} = \angle A \\ {\color{red}x} + A &=& 90^{\circ} \\ {\color{red}x} &=& 90^{\circ} - A \quad | \quad A = 68^{\circ} \\ {\color{red}x} &=& 90^{\circ} - 68^{\circ} \\ {\color{red}x} &=& 22^{\circ} \\ \hline \end{array}$

$\text{$\angle OBC$ in degrees is $22^{\circ}$}$