Question

在三角形ABC中,已知a/√3cosA=c/sinC.(1 求A的大小 (2若a=6,求b+c取值范围

Answer 1

利用正弦定理a/sinA=c/sinC
∵ a/√3cosA=c/sinC
∴ sinA=√3cosA
∴ tanA=√3
（1）A=π/3
（2）a=6
∵ 正弦定理a/sinA=b/sinB=c/sinC
∴ b/sinB=c/sinC=6/(√3/2)=4√3
∴ b+c
=4√3(sinB+sinC)
=4√3[sinB+sin(120°-B)]
=4√3(sinB+sin120°cosB-cos120°sinB)
=4√3[(3/2)sinB+(√3/2)cosB]
=12[sinB*(√3/2)+cosB*(1/2)]
=12(sinBcos30°+cosBsin30°)
=12sin(B+30°)
∵ 0