3、(1)设CD⊥AB，垂足为D，则sinA=DC/AC=DC/b,即DC= b sinAcosB=DB/BC=DB/a,即DB=a cosB又tanB=DC/DB,即tanB= b sinA/ a cosB=√3，B=π/3(2)sinC=sin(π-B-A)=sin(π2/3-A)=sin(π2/3)cosA-cos(π2/3)sinA=√3/2cosA+1/2sinA又有sinC=2sinA,则有√3/2cosA+1/2sinA=2sinA即tanA=1/√3,A=π/6由bsinA= √3acosB得a=bsinA/(√3 cosB)=3sin(π/6) /[√3 cos(π/3)]= √3,∵A+B...
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