Question

三角形ABC中，角A,B,C所对的边分别为a,b,c 且a²=c²+b² +(根号3)ab

(2)设a =根号3,S为三角形ABC的面织，求S+3cosBcosC的最大值，并指出此时B的值
(1)求A

Answer 1

a²=c²+b² +√3ab
cosA=(c²+b² -a²)/2ab=-√3ab/2ab=-√3/2
A=2π/3
B+C=π/3
2R=a/sinA=√3/√3/2=2
2R=a/sinA=b/sinB=c/sinC
S=1/2bcsin2π/3=1/2*4sinBsinC2
题有错，是S+3cosBcosC，还是S+√3cosBcosC
S+3cosBcosC=√3sinBsinC+√3cosBcosC
=√3cos(B-C)
=√3cos(π/3-2C)
0<C<π/3

追问

cosA=(c²+b² -a²)/2ab错了，应该是cosA=(c²+b² -a²)/2bc