1
AB·AC=|AB|*|AC|cosA,而：S=(1/2)|AB|*|AC|sinA,故：|AB|*|AC|=2S/sinA
故：AB·AC=(2S/sinA)*cosA=S,即：tanA=sinA/cosA=2
故：tan2A=2tanA/(1-tanA^2)=4/(-3)=-4/3
2
tanA=2>sqrt(3),故：π/3