设a,b,c分别是三角形ABC中角A,B,C所对的边长,则直线sinA*x+ay+c=0与bx-sinB*y+sinC=0的位置关系是()因为a\sinA=b\SinB所以bSinA=aSinB所以A1A2+B1B2=bsinA-asinB=0所以垂直B 垂直我知道答案是这个,但我还有几个
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![设a,b,c分别是三角形ABC中角A,B,C所对的边长,则直线sinA*x+ay+c=0与bx-sinB*y+sinC=0的位置关系是()因为a\sinA=b\SinB所以bSinA=aSinB所以A1A2+B1B2=bsinA-asinB=0所以垂直B 垂直我知道答案是这个,但我还有几个](/uploads/image/z/14543084-20-4.jpg?t=%E8%AE%BEa%2Cb%2Cc%E5%88%86%E5%88%AB%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E9%95%BF%2C%E5%88%99%E7%9B%B4%E7%BA%BFsinA%2Ax%2Bay%2Bc%3D0%E4%B8%8Ebx-sinB%2Ay%2BsinC%3D0%E7%9A%84%E4%BD%8D%E7%BD%AE%E5%85%B3%E7%B3%BB%E6%98%AF%EF%BC%88%EF%BC%89%E5%9B%A0%E4%B8%BAa%5CsinA%3Db%5CSinB%E6%89%80%E4%BB%A5bSinA%3DaSinB%E6%89%80%E4%BB%A5A1A2%2BB1B2%3DbsinA-asinB%3D0%E6%89%80%E4%BB%A5%E5%9E%82%E7%9B%B4B+%E5%9E%82%E7%9B%B4%E6%88%91%E7%9F%A5%E9%81%93%E7%AD%94%E6%A1%88%E6%98%AF%E8%BF%99%E4%B8%AA%2C%E4%BD%86%E6%88%91%E8%BF%98%E6%9C%89%E5%87%A0%E4%B8%AA)
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