(1)由正弦定理,
(3a-c)/b＝(3sinA-sinC)/sinB=cosC/cosB
所以,cosCsinB=3sinAcosB-sinCcosB
所以cosCsinB+sinCcosB=3sinAcosB
又因为cosCsinB+sinCcosB=sin(B+C)=sinA (A+B+C=180)
所以sinA=3sinAcosB
所以cosB=1/3
所以sinB=2根号2/3
(2)(2)过B做BE⊥AC于E,则CE=2√2,
cosC=CD/b=CE/a
(2/3c)/4√2=2√2/c
c=2√6
BE=4
S=1/2*4*4√2=8√2