在△ABC中,若A=60,a=根号3 则(a+b-c)/(sin A+sin B+sin C)等于?

问题描述:

在△ABC中,若A=60,a=根号3 则(a+b-c)/(sin A+sin B+sin C)等于?
怎么处理(a+b-c)/(sin A+sin B+sin C)?
写错了,是(a+b-c)/(sinA+sinB-sinC)

由a/sina=√3/sin60=b/sinb=c/sinc=2
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