2010 AMC 10B Problem 6

Consider the following diagram: We have that $$\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$$ Since $\triangle AOC$ is isosceles, we have that $$\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$$ Since $\angle CAB = \angle CAO,$ we have that $\angle CAB = 25^{\circ}.$ Thus, B is the correct answer.