Maybe some other way to do this.....but....see the following image :
![](/img/upload/1d96c51c1b945c7b561bbfe2/capture889.png)
We have a circle with a radius of 13 [ any radius would actually be Ok ]
Let a have the endpoints (13,0) and (13√3/2, 13/2)
So the length of a is √ [ ( 13√3/2 - 13)^2 + (13/2)^2 [ = 13/2 ( √6 - √2)
Let b have the endoints ( 13√3/2, 13/2) and ( -13/2, 13√3/2)
So the length of b is √ [ (-13/2 - 13√3/2)^2 + (13/2 - 13√3/2)^2 ] = 13√ 2
So
a + b = (13/2) [ √2 + √6 ] = c = 6.5 [ √2 + √6 ]
So.......using the Law of Cosines
( 6.5 [ √2 +√6] )^2 = 13^2 + 13^2 - 2 (13)(13) cos (degree of arc subtended by c )
Call the degree of arc subtended by c = angle A
So
( 6.5 [ √2 +√6])^2 = 338 - 338 cos (A)
[ ( 6.5 [ √2 +√6])^2 - 338 ]/ -338 = cos A
Using the arccos to find A we have
arccos ( [ ( 6.5 [ √2 +√6])^2 - 338 ]/ -338 ) = A = 150°
So..... chord c subtends an arc of 150°
![cool cool](/img/emoticons/smiley-cool.gif)