Side Length of a 17-gonDate: 03/03/2002 at 17:43:44 From: Jenny Subject: Side length of a 17-gon I have constructed a 17-gon inscribed in a circle of radius 1. What is the length of the side of the 17-gon to 6 decimals? Thank you. Date: 03/04/2002 at 12:28:19 From: Doctor Paul Subject: Re: Side length of a 17-gon From the center of your 17-gon, draw two straight line to adjacent vertices. Each of these lines is a radius of the circle that can be circumscribed about the 17-gon, so each of these lines has length 1. Thus you have an isosceles triangle. You know that the length of the smaller angle is 360/17 degrees. This follows from the fact that there are 360 degrees in a circle and from making the following observation: If you were to draw a line from the center of the 17-gon to every vertex, you would have 17 equal angles in the center of the 17-gon. Clearly the sum of these 17 angles is 360 degrees, so each one must measure 360/17 degrees. Thus you have an isosceles triangle with smaller angle 360/17 degrees and the length of the two congruent sides is 1. You want to solve for the length of the side opposite the smaller angle. You can use the law of cosines to solve for the length of the missing side: cos(A) = (c^2 + b^2 - a^2) / (2bc) we want to solve for 'a' rearranging gives: a^2 = c^2 + b^2 - [2bc * cos(A)] = 1^2 + 1^2 - [2*1*1 * cos(360/17)] = .135055541191 taking the square root gives: a = .367499035633 I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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