在△ABC中,已知AB=3,AC=6,BC=7,AD是 ∠BAC平分线.求(1)DC=2BD,(2)求向量AB,向量DC的值.(ABC是锐角三角形,D在BC上.)
问题描述:
在△ABC中,已知AB=3,AC=6,BC=7,AD是 ∠BAC平分线.求(1)DC=2BD,(2)求向量AB,向量DC的值.(ABC是锐角三角形,D在BC上.)
答
1.因为AB/sin角ABD=BD/sin角BADAC/sin角CAD=CD/sin角CAD又角BAD=角CAD sin角ABD=sin(180-角CAD)=sin角CAD得CD/BD=AC/AB=6/3=2得CD=2BD2.向量AB点乘向量DC=|AB|*|CD|*cos(180-角B)=|AB|*|CD|*cos...