Question

在三角形ABC中tanA:tanB:tanC=1:2:3求AC/AB

Answer 1

A 是锐角

设tanA=m (m>0)

tanB=2m

tanC=tan[π-(A+B)]

=-tan(A+B)

=-(m+2m)/(1-2m²)=3m

所以 1-2m²=-1

m=1

所以 tanA=1,tanB=2,tanC=3

sinA=√2/2,sinB=2/√5,sinC=3/√10

AC/AB=sinB/sinC=(2/√5)/(3/√10)=(2/√5)*√10/3=2√2/3