Question

在△ABC中，若A=60°a=根号3，则a+b-c\sinA+sinB-sinC等于？

在△ABC中，若A=60°a=根号3，则a+b-c\sinA+sinB-sinC等于？ A2 B0.5 C根号3 D根号3\2

Answer 1

A由正弦定理得 a/sinA=b/sinB=c/sinC=根号3/sin60=2 设a/sinA=b/sinB=c/sinC=m则a+b-c/sinA+sinB-sinC=m(sinA+sinB-sinC)/sinA+sinB-sinC=m=2

Answer 2

由a/sina=√3/sin60=b/sinb=c/sinc=2 又因为A=60°
sinA=a/2
(a+b-c)/(sinA+sinB-sinC)
=(a+b-c)/(a/2+b/2-c/2)
=(a+b-c)/[(a+b-c)/2]
=2(a+b-c)/(a+b-c)
=2