△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B

△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=a
sin(∠ABC/2)=根号3/3 cos(∠ABC/2)=根号6/3
sin∠ABC=2sin(∠ABC/2)cos(∠ABC/2)=2根号2/3
cos∠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)
两式相加4x^2+16/3-4+2x^2+32/3-2BC^2=0
3x^2=BC^2-6 (1)
又由余弦定理AC^2=AB^2+BC^2-2AB*BCcos∠ABC
9x^2=4+BC^2-4BC*1/3
27x^2=12+3BC^2-4BC (2)
联立(1)(2) BC=15/2
三角形DBC的面积=(1/2)AB*BCsin∠ABC
=(1/2)*2*(15/2)*2√2/3
=5√2
再问： 答案BC得3额
再答： 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=0 3m^2=BC^2-6 (1) 又由余弦定理AC^2=AB^2+BC^2-2AB*BCcos∠ABC 9m^2=4+BC^2-4BC*1/3 27m^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