xueliang555
幼苗
共回答了21个问题采纳率:85.7% 举报
因C=π/3,所以 A+B=2π/3,A=2π/3-B
sinB=2SinA=2Sin(2π/3-B)=2(sin(2π/3)cosB-cos(2π/3)sinB)=根号3*cosB+sinB
根号3*cosB=0,cosB=0,得B=π/2,从而A=π/6,知三角形ABC为直角三角形.
由正弦定理
c/sinC=a/sinA=b/sinb,即2/sin(π/3)=a/sin(π/6)=b/sin(π/2)
得a=2/(根号3),b=4/(根号3)
所以三角形ABC面积为S=1/2ac=1/2×2×2/(根号3)=2/(根号3)
1年前
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