在三角形ABC中,(a-b)(sinA+sinB)=c(sinB+sinc),(1)求A;(2)若周长为15,且7(sinB+sinC)=8sinA,求面积S

问题描述:

在三角形ABC中,(a-b)(sinA+sinB)=c(sinB+sinc),(1)求A;(2)若周长为15,且7(sinB+sinC)=8sinA,求面积S

(1)(a-b)(a/2r+b/2r)=c(b/2r+c/2r),a²-b²=bc+c²,余弦定理:a²=b²+c²-2bccosA,得:cosA=-1/2,A=120°;
(2)7(sinB+sinC)=8sinA,7b+7c=8a,a+b+c=15,得:a=7,b+c=8,则,b²+bc+c²=49,bc=15,b=3或b=5,c=5或c=3;S=√[15/2(1/2)(9/2)(5/2)]=15√3/4.