smc

Because the triangle is inscribed, the sum of the minor arcs equals $360^{\circ }$. Thus, $$x+75+2x+25+3x-22=360$$ $$6x+78=360$$ Solving this yields $x=47$, so the inscribed angles are $122^{\circ }$, $99^{\circ }$, and $119^{\circ }$. Noting that an angle of $△ABC$ is half of its corresponding inscribed angle, so the angles of $△ABC$ are $59.5^{\circ }$, $49.5^{\circ }$, and $\boxed{61^{\circ }}$.