解:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3tan^2[(B+C)/2]+sin^2 (A/2)=tan^2(π-A)/2+sin^2(A/2)=cot^2(A/2)+sin^2(A/2)=(cos^2(A/2)/sin^2(A/2)+sin^2(A/2)=[cos^2(A/2)+sin^2(A/2)*sin^2(A/2)]/sin^2(A/2)=[cos^2(A/2)+(1-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)=[cos^2(A/2)+sin^2(A/2)-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)=[1-(1/4)sin^2A]/[(1-cosA)/2]=[1-(1/4)2√2/3]/[(1-1/3)/2]=(6-√2)/2(2)S=(1/2)bcsinA=(1/2)bc*2√2/3=√2所以bc=3根据余弦定理a^2-b^2-c^2+2bccosA=0,即4-b^2-c^2+2bc*1/3=0解得:b=c=√3