B=π/， a/8)=sin(3π/4)=b/sin(π/2absinC =1/4) =√2/8) b=2sin(π/8;2×√2×2sin(π/, C=π-A-B=5π/, sinC=sin(5π/2×sin(π/8)=cos(π/8) =√2/解;2×√2/sinA=b/4： A=2B=π/sinB √2/8;8)×cos(π/8) 根据正弦定理;2 =1/sin(π/8) S△ABC=1/
