如何从三个点计算的角度θ [英] How to calculate an angle from three points?

问题描述

可以说，你有这样的：

P1 = (x=2, y=50)
P2 = (x=9, y=40)
P3 = (x=5, y=20)

假设 P1 是一个圆的中心点。它始终是相同的。 我想这是由由 P2 和 P3 ，或者换句话说，就是旁边的<角角code> P1 。内角被precise。它将永远是锐角，所以小于-90度

Assume that P1 is the center point of a circle. It is always the same. I want the angle that is made up by P2 and P3, or in other words the angle that is next to P1. The inner angle to be precise. It will always be an acute angle, so less than -90 degrees.

我想：人，这是简单的几何运算。但我已经找了现在的约6小时，配方，只有找人谈论复杂NASA这样的东西ARCCOS和向量数量积的东西。我的头感觉就像是在冰箱中。

I thought: Man, that's simple geometry math. But I have looked for a formula for around 6 hours now, and only find people talking about complicated NASA stuff like arccos and vector scalar product stuff. My head feels like it's in a fridge.

一些数学大师在这里，觉得这是一个简单的问题？我不认为这里的编程语言重要，但对于那些谁认为它：JAVA和Objective-C。我需要两个，但没有它标记为这些。

Some math gurus here that think this is a simple problem? I don't think the programming language matters here, but for those who think it does: java and objective-c. I need it for both, but haven't tagged it for these.

推荐答案

如果你指的是角度P1是那么这应该工作的顶点：

If you mean the angle that P1 is the vertex of then this should work:

阿尔科斯 （（P ₁₂ ²   + P ₁₃ ² - P ₂₃ ² ）/（2 *   P ₁₂ * P ₁₃ ））

arcos((P₁₂² + P₁₃² - P₂₃²) / (2 * P₁₂ * P₁₃))

其中，P ₁₂ 是从P1段P2，由

where P₁₂ is the length of the segment from P1 to P2, calculated by

开方（（P1 <子> X -   P2 <子> X ） ² +   （P1 <子>是 -   P2 <子>是） ² ）

sqrt((P1_x - P2_x)² + (P1_y - P2_y)²)

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