\(\displaystyle \tan\left(30^\circ\right) = \frac{1}{\sqrt3} = \frac{b/c - 1}{1 + b/c} = \tan\left(B - 45^\circ\right) \implies B = 75^\circ\)

Nicely done. As inspired,

\(\displaystyle b-c=1\)

\(\displaystyle b+c=\sqrt3\)

\(\displaystyle 2b=\sqrt3+1\)

\(\displaystyle b=\frac{\sqrt3+1}{2}\)

Similarly,

\(\displaystyle c=\frac{\sqrt3-1}{2}\)

Using the Pythagorean theorem,

\(\displaystyle a=\sqrt2\)

Then,

\(\displaystyle \sqrt2\sin(B)=\frac{\sqrt3+1}{2}\)

\(\displaystyle \sin(B)=\frac{\sqrt3}{2}\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\frac12=\sin(60^\circ+45^\circ)=\sin(105^\circ)=\sin(75^\circ)\)

As triangle $ABC$ is right, $B=75^\circ$.