Geometry

I am assuming that we have a triangle ABC even though not posted in the question. If that is so, then the first step is to identify a few extra triangles we can draw in. 

I drew an altitude from point C down to AB, and called the point of intersection D. 

Angle A is 45 degrees, so triangle ADC is a right isosceles triangle with hypotenuse AC. 

Angle B is 30 degrees, so triangle BCD is a 30-60-90 triangle with hypotenuse BC.

If those are the cases, then DC is ${\sqrt{6}\over2}$. We also know that AD = DC, so AD = ${\sqrt{6}\over2}$. BD is $\sqrt{3}$ x DC, so BD is $3\sqrt{2}\over2$.

What we also do know is that AB = BD + AD = ${\sqrt{6}\over2} + {3\sqrt{2}\over2}$.

Thus, $AB = {\sqrt{6} + 3\sqrt{2}\over2}$.

PRYAP

Using sin law

(first note that angle C = 180 - 45-30 = 105 degrees)

sin A / sqrt 6 =  sin 105 / AB

AB = sqrt 6   sin 105 / sin 45