如图所示,直角坐标系中有一矩形OADB,OA与x轴正半轴夹角为30°,OA=2,OB=1,对角线AB、OD相交于点C.求ABCD各点的坐标.三角函数没有学
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![如图所示,直角坐标系中有一矩形OADB,OA与x轴正半轴夹角为30°,OA=2,OB=1,对角线AB、OD相交于点C.求ABCD各点的坐标.三角函数没有学](/uploads/image/z/13916764-28-4.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E7%9B%B4%E8%A7%92%E5%9D%90%E6%A0%87%E7%B3%BB%E4%B8%AD%E6%9C%89%E4%B8%80%E7%9F%A9%E5%BD%A2OADB%2COA%E4%B8%8Ex%E8%BD%B4%E6%AD%A3%E5%8D%8A%E8%BD%B4%E5%A4%B9%E8%A7%92%E4%B8%BA30%C2%B0%2COA%3D2%2COB%3D1%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAB%E3%80%81OD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9C.%E6%B1%82ABCD%E5%90%84%E7%82%B9%E7%9A%84%E5%9D%90%E6%A0%87.%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E6%B2%A1%E6%9C%89%E5%AD%A6)
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