(1)sinC/sinBcosA=2c/b → b*sinC-2c*sinBcosA=0
又正弦定理:b/sinB=c/cinC → c*sinB=b*sinC 代入 b*sinC-2c*sinBcosA=0 有
b*sinC(1-2cosA)=0 → 1-2cosA=0 → cosA=1/2 → A=60°
(2)S△=1/2*bcsinA=2√3 → 1/2*4*c*√3/2=2√3 → c=2
余弦定理:a²=b²+c²-2bccosA=16+4-16*1/2=12
a=2√3 ~~ 哟,发现 ∠B=90°