已知:RT三角形ABC中,∠C=90°,AC=4,cotB=根号3,MN=2,QM垂直于AB

已知:RT三角形ABC中,∠C=90°,AC=4,cotB=根号3,MN=2,QM垂直于AB,PN平行于QM,设AM=x,MNPO的面积为y1)当X=5/6时,球PB的长... 已知:RT三角形ABC中,∠C=90°,AC=4,cotB=根号3,MN=2,QM垂直于AB,PN平行于QM,设AM=x,MNPO的面积为y
1) 当X=5/6时,球PB的长
2)求y关于x 的函数解析式,并写出X的取值范围。
3)三角形PCQ与三角形QMA能相似吗?若能,请写出X的值,若不能,请说明理由
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1092465387
2011-11-08
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cotB=根3,所以B=30°,所以AB=AC/sin30°=8,所以BN=AB-AN=31/6,所以PB=BN/cos30,=31根3/9
QM=tan60X,PN=(8-X-2)tan30,所以带入梯形面积公式,y=2根3/3X+2根3,0≤X≤6
假设其相似,则Q一定在AC边上,所以∠A=∠CPQ=60°,所以AQ=2X,QC=4-2X,过Q作QE垂直PN,则MN=QE=2,且,∠PQE=30°,所以PQ=4/根3,所以由三角形相似,PQ/AQ=QC/QM
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旷夏侯容60
2011-11-01 · 超过17用户采纳过TA的回答
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第一问,cotB=根3,所以B=30°,所以AB=AC/sin30°=8,所以BN=AB-AN=31/6,所以PB=BN/cos30,=31根3/9
第二问,QM=tan60X,PN=(8-X-2)tan30,所以带入梯形面积公式,y=2根3/3X+2根3,0≤X≤6
第三问,假设其相似,则Q一定在AC边上,所以∠A=∠CPQ=60°,所以AQ=2X,QC=4-2X,过Q作QE垂直PN,则MN=QE=2,且,∠PQE=30°,所以PQ=4/根3,所以由三角形相似,PQ/AQ=QC/QM,所以带入数据我自己随便算了下,两个X都是负值,应该不存在这样的X吧,具体应该是这样写,答案你在自己算算吧,我不确定~谢谢哈~打字好麻烦,打了好长时间呢~
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弃战遗3132
2011-11-12 · TA获得超过7.7万个赞
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第一问,cotB=根3,所以B=30°,所以AB=AC/sin30°=8,所以BN=AB-AN=31/6,所以PB=BN/cos30,=31根3/9
第二问,QM=tan60X,PN=(8-X-2)tan30,所以带入梯形面积公式,y=2根3/3X+2根3,0≤X≤6
第三问,假设其相似,则Q一定在AC边上,所以∠A=∠CPQ=60°,所以AQ=2X,QC=4-2X,过Q作QE垂直PN,则MN=QE=2,且,∠PQE=30°,所以PQ=4/根3,所以由三角形相似,PQ/AQ=QC/QM,所以带入数据我自己随便算了下,两个X都是负值,应该不存在这样的X吧,具体应该是这样写,答案你在自己算算吧,我不确定~谢谢哈~打字好麻烦,打了好长时间呢~
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