Question 53918
<pre><font size = 2><b>When graphed using polar coordinates, the center of a regular nonagon is at the
origin and one vertex is at (6,0 degrees) or (6,0 radians). Find the polar
coordiantes of the other vertices in both degrees and radians

The vertices of a nonagon are the enpoints of the "spokes" of a "9-spoke wheel",
each spoke measuring 6 units. 

We divide 360° into 9 parts, 360°/9 = 40°.  So the "spokes" are 40° apart, so
the vertices have as their second coordinates

0°, 40°, 80°, 120°, 160°, 200°, 240°, 280°, 320°

Each "spoke" is 6 units long, so the polar coordinates for the 9 vertices are

(6,0°), (6,40°), (6,80°), (6,120°), (6,160°), (6,200°), (6,240°), (6,280°), (6,320°)

In radians, 40° = 40(<font face = "symbol">p</font>/180) = 2<font face = "symbol">p</font>/9

So the polar coordinates in radians are:

(6,0), (6,2<font face = "symbol">p</font>/9), (6,4<font face = "symbol">p</font>/9), (6,6<font face = "symbol">p</font>/9), (6,8<font face = "symbol">p</font>/9), (6,10<font face = "symbol">p</font>/9), (6,12<font face = "symbol">p</font>/9), (6,14<font face = "symbol">p</font>/9), (6,16<font face = "symbol">p</font>/9)

Some of those fractions will reduce, so, after reducing those that will reduce
the vertices of the nonagon in radians are:

(6,0), (6,2<font face = "symbol">p</font>/9), (6,4<font face = "symbol">p</font>/9), (6,2<font face = "symbol">p</font>/3), (6,8<font face = "symbol">p</font>/9), (6,10<font face = "symbol">p</font>/9), (6,4<font face = "symbol">p</font>/3), (6,14<font face = "symbol">p</font>/9), (6,16<font face = "symbol">p</font>/9)

Edwin</pre>