已知△ABC中,∠A=60°,S△ABC=根号3,a+b-c/sinA+sinB-sinC=2根号39/3,求b
问题描述：
已知△ABC中,∠A=60°,S△ABC=根号3,a+b-c/sinA+sinB-sinC=2根号39/3,求b
答
(a+b-c)/(sinA+sinB-sinC)=K(sinA+sinB-sinC)/(sinA+sinB-sinC)=k,则a/sinA=k(正弦定理）,即a=SINA*k=SQR(13),又三角形面积等于=1/2bcsinA=SQR(3),则bc=4,又a^2=b^2+c^2-2bcCOSA,则带入得13=b^2+16/b^2-4,解得：b=4,...