在三角形ABC中,边a,b,c所对的角A,B,C组成一个公差为α的等差数列.（1）:a=2,c=3,求tanα的值.（2

A=B-d,B=B,C=B+d
A+B+C=π
B=π/3
A=π/3-d,C=π/3+d
a/sinA=c/sinC
csinA=asinC
3sin(π/3-d)=2sin(π/3+d),
3√3/2cosd-3/2sind=√3cosd-sind
sind=√3cosd
tand=√3
2)a＋c=λb
λ=(a+c)/b=(sinA+sinC)/sinB=[sin(π/3-d)+sin(π/3+d)]/sinπ/3
=2√3/3[√3/2cosd-1/2sind+√3/2cosd+1/2sind]
=2√3/3*√3cosd
=2cosd
三角形ABC为锐角三角形,A,B,C组成一个公差为α的等差数列
d在[0,π/6)
cosd在(√3/2,1]
2cosd在(√3,2]
λ在(√3,2]