在△ABC中,内角A,B,C所对边的长分别是a,b,c,cosC+(cosA-√3sinA)cosB=0

2025-03-07 10:04:48
推荐回答(1个)
回答1:

(1)
cosC=cos(180-A-B)=-cos(A+B)=-cosAcosB+sinAsinB;又
cosC+(cosA-√3sinA)cosB=0即cosAcosB-√3sinAcosB=cosAcosB-sinAsinB;
tanB=√3,又B是内角,B=60
(2)s=1/2absinC=5√3,b=4;
又a/sinA=b/sinB所以a=bsinA/sinB,s=1/2absinC=1/2b²sinAsinC/sinB=5√3
代入sinB=√3/2,b=4得:sinAsinC=15/16