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[几何] 解三角形b:c

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realnumber posted 2018-5-25 20:30 |Read mode
三角形ABC中,角A,B,C所对边分别为a,b,c,且$b^2\cos{A}=5c^2-a^2$.
求b:c (答案2,如果不看答案,或者不知道能求出来, )

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isee posted 2018-5-25 21:29
Last edited by isee 2018-5-26 00:35回复 1# realnumber


第一直觉,化边,a,b,c关系总是有的。



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动笔算了下,b^3与10c^3,肯定拆成8c^3与2c^3

由于提前看到了结果,顺手就分解因式Ok了
依然饭特稀 posted 2018-5-25 22:50
可以考虑取c=1,把cosA表示出来,再用余弦定理,对照系数

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游客 posted 2018-5-28 10:25
题目的意思是b:c跟a的取值无关,类似与函数图像过定点的问题。

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