在三角形ABC中,cos二分之B的平方等于2c分之b＋c 则三角形ABC的形状为?
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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.由面积公式,得:S=0.5*b*c*sinA 代入已知:根号3=0.5*1*c*sin60度.根号3=0.5*1*c*(根号3)/2 得:c=4.由余弦定理:a^2=b^2+c^2-2*b*c*cosA 得:a^2=1+16-2*1*4*(1/2)=13.即:a=根号13.由正弦定理:a/sinA=b/sinB=c/sinC,由比的性质,有a/sinA=b/sinB=c/sinC= =(a+b+c)/(sinA+sinB+sinC).而已知:a/sinA=根号13/[(根号3)/2] =2*(根号39)/3 即:(a+b+c)/(sinA+sinB+sinC)=2*(根号39)/3.
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