∠A+∠B+∠C=π故cos(2B+C)=cos[π-(A-B)]=-cos(A-B)于是-cos(A-B)2sinAsinBcosAcosB+sinAsinB>2sinAsinBcosAcosB-sinAsinB>0cos(A+B)>0由于∠A+∠B+∠C=π故A+B=π-Ccos(π-C)>0cosCπ/2所以三角形是钝角三角形...