已知bcosC+ccosB=2/3根号3asinA
由正弦定理化为角的形式
sinBcosC+sinCcosB=(2/3)√3*sin²A
sin(B+C)=(2/3)√3*sin²A
sinA=(2/3)√3*sin²A
即sinA=√3/2
因是锐角三角形ABC
所以A=60°
B=180°-A-C=180°-60°-C=120°-C
由正弦定理(b-c)/a=(sinB-sinC)/sinA
=2cos[(B+C)/2]sin[(B-C)/2]/(√3/2)
=(4/3)√3*cos(90°-A/2)sin[(120°-C-C)/2]
=(4/3)√3*cos60°sin[(120°-2C)/2]
=(2/3)√3*sin(60°-C)
因C为锐角,即0