Question

在三角形ABC中,cos平方A/2=（b+c）/2c,

Answer 1

由正弦定理:b/2R=sinB,c/2R=sinC 所以(b+c)/2c =[(2RsinB)+(2RsinC)]/[2(2RsinC)] =(sinB+sinC)/2sinC 所以:cos^2(A/2)=(sinB+sinC)/2sinC (cosA+1)/2=(sinB+sinC)/2sinC (cosA+1)sinC=sinB+sinC cosAsinC=sinB =sin(π-A-C) =sin(A+C) =sinAcosC+cosAsinC 所以sinAcosC=0 因为A是三角形内角,所以sinA>0 故cosC=0 C=90° 所以三角形是直角三角形