Question

在△ABC中，若∠A:∠B:∠C=4:2:1,a,b,c分别为∠A，∠B，∠C的对边，求证：1/a+1/b=1/c

Answer 1

设C=X 则B=2X， A=4X 由a/sinA=b/sinB/=c/sinC=2R (R为三角形外接圆半径) 得1/a=1/(2RsinA) 1/b=1/(2RsinB) 1/c=1/(2RsinC) 1/c=1/(2RsinX) ① 1/a+1/b=1/2R*(1/sin4X+1/sin2X) =1/2R*[(sin4X+sin2X)/(sin4x*sin2X)] =1/2R*[(2sin3X*cosx)/(2sinX*cosX*sin4X)] =1/2R*(sin3X)/(sinX*sin4X) ∵ A+B+C=180° ∴ 4X+2X=X=180° 即4X=180°-3X sin4X=sin(180°-3X) =sin3X 所以1/a+1/b=1/2R*(sin3X)/(sinX*sin3X) =1/(2RsinX) ② 由①②得 1/a+1/b=1/c