Angle B=48+65=113°.
AC=√(AB²+BC²-2AB.BCcos113)=√(650²+810²-2*650*810cos113)=
1220.67 miles approx. (Cosine rule)
Let θ be the bearing of C, then:
sin(θ-48)/810=sin113/1220.67, (sine rule)
sin(θ-48)=810sin113/1220.67=0.6108 approx.
So θ=48+37.65=85.65° (approx). The bearing is N85.65°E.