f(x)=√3sin^2x+sinxcosx-√3/2
=-√3/2*cos2x+1/2*sin2x
=sin(2x-π/3),
f(α)=sin(2α-π/3)=4/5,
α∈(0,π/2),
2α-π/3∈(-π/3,2π/3),
∴cos(2α-π/3)=土3/5,
∴cos2α=cos(2α-π/3+π/3)=土3/5*1/2-4/5*√3/2=(土3-4√3)/10.
A∴2A-π/3=π/6,2B-π/3=5π/6,
∴A=π/4,B=7π/12,C=π/6,
∴BC/AB=sinA/sinC=√2.
图片清楚
这种做法 是最好的,可以给最佳了