I used to despise constructions back in school

Let's go back to school and treat this as a construction problem ;)

We can draw the line segment AB = 95, and we draw a ray R emanating from A, making an angle of $30^o$ with AB, there C will be placed. The distance between R and B is $\sin(30^o)\times95=47.5$ , so that the circle about B with radius 55 will intersect R twice (since $55<95$ ), giving us our $\boxed{2}$ solutions C.

Another possible approach is to use Law of Cosines:

$BC^2=AB^2+AC^2-2AB\cdot AC \cos(\angle A) \\ 55^2=95^2+AC^2-2(95)(AC)\cos(30^\circ) \\ AC^2-95\sqrt{3}AC+6000=0$

If we compute the discriminant of that quadratic we get:

$\Delta=(-95\sqrt{3})^2-4(1)(6000)=3075$

We got that $\Delta>0$ , so there are $\boxed{2}$ distinct real solutions for $AC$ , and by Descartes Rule of Signs, both are positive.