*flashbacks to honors trig in 9th grade*
...Anyway. We're given that the bearing from A to C is [tex]330^\circ,[/tex] so that means that the counterclockwise angle from the north to C is [tex]30^\circ.[/tex] Therefore, [tex]\angle BAC=50^\circ+30^\circ=80^\circ.[/tex]
Now, we apply the law of cosines:
[tex]BC^2&=AC^2+BC^2-2(AC)(BC)\cos\angle BAC[/tex]
[tex]BC^2&=65^2+135^2-2(65)(135)\cos 80^\circ=22450-17550\cos 80^\circ[/tex]
[tex]\boxed{BC\approx 139\text{ km}}.[/tex]
