在△ABC中,∵tanB=1/2,tanC=-2.
∴sinB=1/√5,cosB=2/√5 ;sinC=2/√5,cosC=-1/√5
∴sinA=sin(B+C)
=sinBcosC+cosBsinC=(1/√5)*(-1/√5)+(2/√5)*(2/√5) =3/5
根据正弦定理有：BC/sinA=AB/sinC=AC/sinB =2R
∴BC/(3/5)=AB/(2/√5)=AC/(1/√5) ===>AB:AC=sinC:sinB=2:1
S△ABC=AB*AC*sinA/2=1
得到：AB=2√(5/3)=2√15/3,BC=√3,AC=√15/3
R=AC/2sinB=√(5/3)/(2*1/√5)=5/(2√3)=5√3/6