三角形abc中,若A=60度,a=根号3,则a+b-c/sinA+sinB-sinC=多少

问题描述:

三角形abc中,若A=60度,a=根号3,则a+b-c/sinA+sinB-sinC=多少

a/sinA=√3/sin60°=√3÷(√3/2)=2
由正弦定理得:a/sinA=b/sinB=c/sinC=2
∴a=2sinA b=2sinB c=2sinC
∴a+b-c/sinA+sinB-sinC
=(2sinA+2sinB-2sinC)/(sinA+sinB-sinC)=2