Question

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

1.a/c的值
2.cotB+cotC的值

Answer 1

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