在△ABC中,a,b,c分别是角A、B、C所对的边,已知
cosB=,
(1)判断△ABC的形状;
(2)若
sinB=,b=3,求△ABC的面积.
(1)∵cosB=a2c,asinA=csinC,∴cosB=sinA2sinC,∴sinA=2cosBsinC,又∵sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosBsinC,∴sinBcosC+cosBsinC=2cosBsinC,∴sinBcosC-cosBsinC=sin(B-C)=0∴在△ABC中B=...