在△ABC中,角A,B,C所对的便是A,B,C,且3bsinC-5csinBcosA=o.[1]求sinA2若tan[A-B]=-2/11,求tanC

Question

在△ABC中,角A,B,C所对的便是A,B,C,且3bsinC-5csinBcosA=o.[1]求sinA 【2】若tan[A-B]=-2/11,求tanC

Answer 1

正弦定理
c/sinC=b/sinB=2R,R为外接圆半径
sinC=c/2R
sinB=b/2R
根据题意
3b×c/2R-5cb/2R×cosA=0
5bccosA=3bc
cosA=3/5
A为锐角
sinA=4/5
sin²A+cos²A=1
tanA=sinA/cosA=4/3
tan2A=2tanA/(1-tan²A)=-24/7
B=180-A-C
tan(A-B)=-2/11
tan[A-(180-A-C)]=-2/11
tan(2A+C)=-2/11
(tan2A+tanC)/(1-tan2AtanC)=-2/11
令tanC=t
(-24/7+t)/(1+24t/7)=-2/11
整理125t=250
t=2
tanC=2