由sin(B+C)+3sin(A+C)cosC=0，得sin(B+C)+3sinBcosC=0，拆开得cosBsinC+4sinBcosC=0，得tanC=-4tanB。tanA=-tan(B+C)=(3tanB)/(1+4tan²B)=3/(1/tanB+4tanB)≤3/4，当tanB=1/2时等号成立
