+1622 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, \(AB = {\sqrt{6} + 3\sqrt{2}\over2}\).
PR
YA
P
+36916 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
