On Tuesday, May 28, 2013 3:24:58 PM UTC-7, JT wrote: > On 28 Maj, 23:24, Ken Pledger <ken.pled...@vuw.ac.nz> wrote: > > > In article <firstname.lastname@example.org>, > > > > > > 130bcd <carl.sundqu...@gmail.com> wrote: > > > > .... 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.... > > > > > > If a regular n-gon is inscribed in a circle of radiu r, then each > > > side of the n-gon has length 2r.sin(pi/n). In your case that means > > > 260.sin(pi/5) or 260.sin(36 degrees). > > > > > > That can actually be expressed in terms of square roots if you like, > > > since sin(36 degrees) = (1/4)sqrt(10 - 2.sqrt(5)). Such expressions > > > involving sqrt(5) underlie the construction of the regular pentagon by > > > straightedge and compasses (Euclid IV.11, or a neater construction by > > > Ptolemy). > > > > > > Ken Pledger. > > > > I see your formula use Pi so i guess your formula can only calculate > > the perimeter to the precision of the given Pi and same would go for > > the area? > > Would it not be beneficial finding a formula using fractions, that > > could calculate the perimeter as well as area exact, without using a > > couple of billions of decimalpoints on Pi? > > > > Except from being accurate it sure would put an ease to the > > calculation machinwise or humanwise.
Nonsense: the formula he gave did not involve pi at the end; it just involved some square roots to get the necessary trig function values.
If we measure angles in radians we get pi in the expression of an ANGLE, but not necessarily in the expressions for trig functions of the angle. If we express angles in degrees, pi never comes into it. What is it about this simple fact that escapes you?