第1个回答 2014-05-08
tanB=√3=sinB/cosB,sin^2B=3cos^2B,sin^2B=3-3sin^2B,sinB=√3/2,cosB=1/2;
cosC=1/3,9cos^2C=1,9sin^2C=8,sinC=2√2/3;
b/sinB=c/sinC,c=bsinC/sinB=3√6*(2√2/3)/(√3/2)=8;
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosBsinC=(√3/2)(1/3)+(1/2)(2√2/3)=(√3+2√2)/6
△ABC的面积=bcsinA/2=[8*3√6*(√3+2√2)/6]/2=6√2+8√3