设三角形ABC的内角A,B,C的对边分别为a,b,c,且A=60度,c=3b,求

a^2=b^2+c^2-2bccosAa^2=(c/3)^2+c^2-2(c/3)c(1/2)a^2=7/9c^2a/c=√7/3由正弦定理得sinA=√7/3sinC,√3/2=√7/3sinC,sinC=3√21/14sinB=√21/14cotB+cotC=sin(B+c)/sinBsinC=(√3/2)/(3√21/14)*(√21/14)=14√3/9...