(1)tanC=(sinA+sinB)/(cosA+cosB)=2sin[(A+B)/2]cos[(A-B)/2]/2cos[(A+B)/2]cos[(A-B)/2]=tan[(A+B)/2]=tan(π-C/2)=1/tan(C/2).
∴tanCtan(C/2)=1.于是設(shè)tan(C/2)=t.則有2t²=1-t².解得t＝±(√3)/3
所以C＝π/3.
sin(B-A)=cosC=-cos(B+A)
所以sinBcosA-sinAcosB+cosBcosA-sinBsinA=0
即(sinB+cosB)cosA=(sinB+cosB)sinA
若sinB+cosB=0.則B=3π/4.若sinB+cosB≠0.則A=π/4
又A+B+C=π.所以(A,B,C)=(π/4,5π/12,π/3)
(2)S=absin(π/3)/2=bcsin(π/4)/2=acsin(5π/12)/2=3+√3
∴a=(√6)c/6.c²sin(5π/12)×(√6)/12＝3+√3
即c²=24.a²=4.故可求得a=2.c=2√6