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标题: 求角度.[@};-] [打印本页]

作者: 寒潭清 时间: 2005-9-3 17:24
标题: 求角度.[@};-]

已知三角形ABC中AB=AC，角A=80度，P为三角形ABC中一点，连接PA，PＢ，ＰＣ，角PAB=70度，角BPC=140度。求其他各角度数。
用初中知识，不能用三角函数

作者: 乱弹 时间: 2005-9-7 00:38
标题: 正三角形的威力 2 +同一法[:L]

设O为与A相对于BC对称的点。在三角形内取点Q，使得BQO成一正三角形。很容易推算出P=Q。然后可以散出各角，比如角APB=70。

作者: husonghu 时间: 2005-9-7 01:31
标题: Good. Very glad to see you again, 乱弹[@};-]

Good. Very glad to see you again, 乱弹

作者: yma16 时间: 2005-9-7 16:34
标题: How to get P=Q?, Thanks - EOM

How to get P=Q?, Thanks - EOM

作者: 乱弹 时间: 2005-9-7 16:56
标题: 回复：Also thank husonghu's [@};-]

Show 角QAB=70度，角BQC=140度。 The point to satisfy this condition is unique. --It is the intersection of a line and an arc.www.ddhw.com

To husonghu: I was away for conferences in Asia, always not very active in every forum though. TCN is a nice place. You have done a nice job.

作者: yma16 时间: 2005-9-7 20:52
标题: How to show 角BQC=140度?-eom

How to show 角BQC=140度?-eom

作者: 乱弹 时间: 2005-9-8 02:40
标题: 回复：How to show 角BQC=140度?-eom

Just note that OB=OC=OQ,

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