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--- math:geometry [2025/12/01 21:35] – ron
+++ math:geometry [2025/12/01 22:05] (current) – ron
@@ Line 6: / Line 6: @@
   * we can easily calculate the side **s** via [General Pythagorean](math:trigonometry)
     * c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> - 2ab * cos(Θ) where c is the side opposite Θ
-    * c = √(a<sup>2</sup> + b<sup>2</sup> - 2ab * cos(Θ))
+  * since the triangle is inscribed within a circle of radius **R** both **a** and **b** = **R**
+    * c = √(R<sup>2</sup> + R<sup>2</sup> - 2RR * cos(Θ))
+      * = √(2R<sup>2</sup> - 2R<sup>2</sup> * cos(Θ))
+      * = R√(2 - 2 * cos(Θ))
 ### Pentagon
   * Θ = 360/5 = 72 degrees (2π/5 radians)
-  * s = √(R<sup>2</sup> + R<sup>2</sup> - 2R<sup>2</sup> * cos(72-deg))
+  * s = R * √(2 - 2 * cos(72-deg)); cos(72) = (-1 + √(5)) / 4 (see [this](https://www.quora.com/How-can-I-get-the-fraction-value-of-cos-72))
-    * = √(2R<sup>2</sup> - 2R<sup>2</sup> * cos(72-deg))
+    * = R * √(2 - 2 * (-1 + √(5)) / 4) = R * √(2 - (-1 + √(5)) / 2) = R * √( (4 + 1 - √5)/2)
-    * = R * √(2 - 2 * cos(72-deg))
+    * = R * √( (5 - √5)/2)
     * ≅ R * 1.17557
 ### Hexagon
   * Θ = 360/6 = 60 degrees
-  * s = √(R<sup>2</sup> + R<sup>2</sup> - 2R<sup>2</sup> * cos(60-deg))
+  * s = R * √(2 - 2 * cos(60-deg)) = R * √(2 - 2 * 0.5) = R * √(1)
-    * = √(2 * R<sup>2</sup> - 2R<sup>2</sup> * 0.5)
-    * = √(2 * R<sup>2</sup> - R<sup>2</sup>)
-    * = √(R<sup>2</sup>)
     * = R