\(\because\) ABCD is a cyclic quadrilateral \(\therefore \angle\)ABC = 180° − \(\angle\)ADC = 180° − 130° = 50° In triangle ABC, \(\angle\)BAC = 90° [angle in a semicircle] \(\therefore \angle\)ACB = 180° − (\(\angle\)BAC + \(\angle\)ABC) = 180° − (90° + 50°) = 180° − 140° = 40°