After b, BD⊥AC is in D.

∠C+∠A+∠B= 180

∠C= 180 -30 - 135 =45

In Rt△BCD.

BD=BCsin∠C=5√2*(√2/2)=5(Rt△BCD is an isosceles right triangle, which can also be solved by Pythagorean theorem).

CD=BD=5

In Rt△ABD

BD/AB=sin∠A=sin30 = 1/2

AB=DB/( 1/2)= 10

ad=√(ab^2-bd^2)=√( 10^2-5^2)=5√3

AC=AD+CD=5√3+5