设△ABC,AD是〈A的坟分线,AD=2,AB=4,AC=3,根据角平分定理,AB/AC=BD/BC=4/3,设BD=4m,CD=3m,在△ABD和△ADC中,根据余弦定理,BD^2=AB^2+AD^2-2*AB*AD*cos(A/2),16m^2=16+4-2*4*2cos(A/2),m^2=5/4-cos(A/2),(1)CD^2=AC^2+AD^2-2*AC*AD*cos(A/2),9m^2=9+4-2*3*2cos(A/2),m^2=13/9-(4/3)cos(A/2),(2)比较(1)和(2)式,cos(A/2)=7/12,cosA=2(cosA/2)^2-1=-23/72,sinA=√[1... |