If O is the center of a circle on which A, B and C are three points; and if $$\angle BOC = 100^\circ$$ then $$\angle ABC + \angle ACB =$$

For ease of determination, make A and B ends of a diameter of a circle. This makes triangle ACB a right angled triangle and hence angle ACB becomes = 90 degrees. 

Now, in isosceles triangle BOC, OB = OC (radii of same circle) and angle BOC = 100 degrees.
Hence, angles OBC and OCB = 40 degrees each. Hence, this also makes angle ABC = 40 degrees. 

Hence, sum of the two required angles = 90+40 = 130 degrees.