The angles form an arithmetic sequence with common difference $\Delta$. [asy] pair A,B,C; A = (0,0); B = (2.25,0); C = (1,1.7); draw(A--B--C--cycle); draw(rightanglemark(B,A,C,2)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); [/asy] Since the sum of the measures of the angles in a triangle is $180^\circ$, we have \[\angle A + \angle B + \angle C = 180^\circ.\]Since $\angle A = 45^\circ$ and the three angles form an arithmetic sequence, we have $\angle B = 45^\circ + \Delta$ and $\angle C = 45^\circ + 2 \Delta$. Substituting these into $\angle A + \angle B + \angle C = 180^\circ$, we get \[45^\circ + (45^\circ + \Delta) + (45^\circ + 2 \Delta) = 180^\circ.\]Solving, we find $\Delta = 30^\circ$, so $\angle C = 45^\circ + 2 \Delta = \boxed{105^\circ}$.
