2025 AMC 10B Problem 5

Since $O$ is the circumcenter, $OA = OB = OC.$ The inscribed angle $\angle B = 130^\circ$ subtends arc $AC,$ and because $B$ is obtuse, the central angle is $\angle AOC = 360^\circ$ $- 2 \cdot 130^\circ = 100^\circ.$ Triangle $OAC$ is isosceles, so $\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ.$ (The lengths $AB$ and $AC$ never enter.) Thus, C is the correct answer.