You don't need to worry about phase angles for this, and you just about never square an angle.
You've worked out ωL in the right way, but the inductor's impedance at 75.96 ohms. (I guess that you didn't put 39 mΩ as 0.039 Ω) That's about the same as the resistance so we are looking at somewhere around 1½ times that, about of 100 Ω
But to work it out correctly,
impedance = √(R²+(ωL)²) = √(71²+75.96²) = 103.98 Ω
(Looks right, about 100 Ω)
so the current is 34000/103.98 = 327.0 Amps
If you did want to work it out with angles, the phase angle is atan (75.96/71) = 46.993 degrees
Then your impedance is either
75.96 / sin (46.993) or 71 / cos (46.993), both of which come out to be 103.98 Ω
(I am so glad this is homework and not a backyard generator project. There are all sorts of warnings that go with 3 MVA systems!)