Since OC is a perpendicular bisector of the chord AB, it also bisects the arc ACB, splitting it into two arcs, AC and CB, each of equal measure. Since the whole arc is 110, mAC = mCB = 55.
(For a more detailed proof of why this fact is true, construct lines AC and BC to make two right triangles, AMC and BMC (where M is the intersection of the radius and the chord, marked by a right angle sign in your diagram). It's generally given in geometry texts that a radius that intersects a chord at a right angle bisects the chord, so we'll take that as a given, so by the defn. of bisectors, AM = BM. Also, AMC is also a right angle. Thus, the two triangles are congruent by SAS. Thus, angle CAB and CBA are congruent inscribed angles, so the arcs they intercept must be congruent as well - thus, mAC = mCB; QED.)
Note that angle 1 intercepts arc CB and is a central angle. Central angles always have the same measure as the arcs they intercept. Since mCB = 55, m<1 = 55.