can anyone figure this out?

Question

I was ask to build A 10' polygon that had 13 sides that were equal in length, what would be the angle of the outside that would create this polygon? and also, what would be the outside length of each side of this polygon?

Answer 1

The measure of the exterior angle is given by 360/13 which is 27.69 degrees.

The measure of each of the 13 central angles is also 27.69 degrees. Now, you have been unspecific about what the 10' is supposed to be (i.e. radius or diameter). Assuming it's the diameter, the radius is 5'. To find the length of a side, we need to consider an isosceles triangle with legs of length 5 and a vertex angle of 27.69 degrees. The two base angles are (180-27.69)/2 = 76.155 degrees. Using law of sines, we get

sin(76.155) / 5 = sin(27.69)/x
x = 5sin(27.69)/sin(76.155)
x = 2.4 feet (approximately)

If you want it in inches, it is 28.71787971... inches. (approximately)

Answer 2

I take it you mean a 10 foot polygon. Whatever the distance, the procedure is still the same.
Draw a 10 foot circle (r = 5 ft.).
Since it has 13 sides, draw the radius coming from the center of the circle 13 times, with an angle of 27∙6923°... apart (ie 360°/13).
Where each radius cuts the circumference of the circle, join these points with straight lines. These lines are the edges of your polygon.
Divide one of the angles between two of the radii at the centre point of the circle and at this angle (27∙6923°.../ 2 = 13∙8461°...) draw a straight line to it's corresponding edge line. This new straight line cuts the edge of the polygon in half.
By trigonometry the half length of the edge of the polygon may be calculated.
Sin Θ = opp. / Hyp.
Sin 27∙6923°... = opp./ 5ft.
Opp. = 5 Sin 13∙8461°...
Opp. = 1.19657..... But this is half the length of the edge.
Length of edge = 1.19657... x 2 = 2∙393156643 ft

Answer 3

it would be 27.69