过D作DE⊥BA交BA的延长线于E,作DF⊥BC交BC于F.
∵∠BAD=120°,∴∠DAE=60°,又DE⊥AE,∴DE=√3AE.
由勾股定理,有:BE^2+DE^2=BD^2,∴(AB+AE)^2+(√3AE)^2=49,
∴(3+AE)^2+3AE^2=49,∴9+6AE+AE^2+3AE^2-49=0,
∴4AE^2+6AE-40=0,∴2AE^2+3AE-20=0,∴(2AE-5)(AE+4)=0,∴AE=5/2.
∴DE=√3AE=(5/2)√3.
∵DE⊥BE、BE⊥BF、DF⊥BF,∴BFDE是矩形,∴BF=DE=(5/2)√3.
∴CF=BC-BF=5√3-(5/2)√3=(5/2)√3,∴BF=CF=(5/2)√3.
∵DF⊥BC、BF=CF,∴CD=BD=7.