Question

在△ABC中,b²+c²-bc=a²,且c/b=1/2+根号3,则tanB=?A.1 B.

Answer 1

由余弦定理,得

cosA＝(b²+c²-a²)/2bc

又,b²+c²-bc=a²

即,b²+c²-a²=bc

所以,cosA＝1/2

A为三角形内角,

所以,A＝π/3

B+C=π-π/3=2π/3

C=2π/3-B

sinC/sinB=sin(2π/3-B)/sinB

=(sin2π/3cosB-cos2π/3sinB)/sinB

=(√3/2)cotB＋(1/2)

又,sinC/sinB=c/b=1/2+√3

所以,

(√3/2)cotB＋(1/2)＝1/2+√3

cotB＝2

tanB＝1/cotB＝1/2

选B.1/2