Subtract the angles

Solution:

$AB$ is in fact diameter for the given circle. So, the segment $BC$ has the length of $r$ and $\angle BCA = 90°$ . Length of $AC$ can be found using Pythagora's theorem and $AC = r\sqrt{3}$ . It is obvious by now, that $\triangle ABC$ is a half of an equilateral triangle. Therefore, $\angle ABC = 60°$ and $\angle BAC = 30°$ . Since $\angle BXC$ and $\angle BAC$ have the common segment $BC$ then $\angle BXC = \angle BAC \vee 180° - \angle BAC$ . Thus, solution is $180° - 2 \times \angle BAC = 120°$ .