
Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 2:34 PM


On Tue, 28 May 2013 08:15:05 0700 (PDT), 130bcd <carl.sundquist@gmail.com> wrote:
>Hello, > >I have unsuccessfully tried to find the perimeter of a pentagon. The "radius" to the vertices is 130, which I tried to convert into 5 triangles but I can't find a formula for determining the length of the 3rd side of an isosceles triangle. > >The two equal length sides would, of course be 130 and the angle between them would be 72.
Try the law of cosines: c^2 = a^2 + b^2  2*a*b*cos(C)
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