2008 AMC 10B Problem 10

Let $O$ be the center and $D$ the midpoint of $AB.$ Then $OD\perp AB$ with $AD=3,$ so $OD=\sqrt{5^2-3^2}=4.$ Since $C$ is the midpoint of the minor arc, $O,$ $D,$ $C$ are collinear and $DC=OC-OD=5-4=1.$ Then $AC=\sqrt{AD^2+DC^2}$ $=\sqrt{3^2+1^2}$ $=\sqrt{10}.$ Thus, the correct answer is A.