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[几何] 三角形旋转问题

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chn-2000 posted 2024-8-10 08:42 |Read mode
Last edited by hbghlyj 2025-4-19 19:02如图,△ABC绕点C逆时针旋转90°后得△DEC,如果点B、D、E在一直线上,且∠BDC=60°,BE=3,那么A、D两点间的距离是?

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isee posted 2024-8-10 11:26
三角形 BCE 是直角三角形且等腰 BC=CE,作斜边 BE 上的高 CF,则 CF=1.5,在直角三角形 CFD 中有已知角 60 度,则直角边 CD可求.
isee=freeMaths@知乎

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original poster chn-2000 posted 2024-8-10 15:32
感谢isee的解答!
我是在△CDE中用正弦定理求出DE的长度,但是这样子就超纲了,初中没有学过正弦定理。
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