△ABC中,sin(∠ABC/2)=根号3/3,AB=2,点D在线段AC上,且AD=2DC,BD=4倍根号3除以3,求BC的长;求三角形DBC的面

设CD=m ∠ADB=asin(∠ABC/2)=根号3/3 cos(∠ABC/2)=根号6/3sin∠ABC=2sin(∠ABC/2)cos(∠ABC/2)=2根号2/3cos∠ABC=1/3由余弦定理 cosa=(4x^2+16/3-4)/(16√3m/3)cos(180°-a)=-cosa=(x^2+16/3-BC^2)/(8√3m/3)两式相...答案BC得3额由余弦定理 cosa=(4m^2+16/3-4)/(16√3m/3)cos(180°-a)=-cosa=(m^2+16/3-BC^2)/(8√3m/3)两式相加4m^2+16/3-4+2m^2+32/3-2BC^2=03m^2=BC^2-6 (1)又由余弦定理AC^2=AB^2+BC^2-2AB*BCcos∠ABC9m^2=4+BC^2-4BC*1/327m^2=12+3BC^2-4BC (2)联立(1)(2)BC=3 (对的，你对)三角形ABC的面积=(1/2)AB*BCsin∠ABC=(1/2)*2*3*2√2/3=2√2 三角形DBC的面积=(1/3)三角形ABC的面积=2√2/3