It's easy to see that $ \angle A=30^o$ satifies the supposion. Assuming there exist $ \angle B'AC\neq 30^o$ and satify the supposion too. I only solve this problem when $ \angle B'AC<30^o$ (*)(another case is similar) Assuming that a point M' satifies the supposion. We have $ \angle MB'B<\angle MBz=\angle ABC<\angle AB'C$ then $ 90^o-\angle MB'B>90^o-\angle AB'C$ $ \Rightarrow \angle MB'y>\angle AB'y=\angle M'B'y$ Therefore B'B' is between B'M and B'y which follow M' is between M and E. $ \Rightarrow \angle B'AC=\angle M'CB'>\angle MCB=\angle BAC$ which is contrary with $ (*)$ So this problem only has one solution: $ \angle A=30^o$

Attachments:
picture13.pdf (4kb)