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Maybe some other way to do this.....but....see  the following image :

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°

For the chord of length a, take the isosceles triangle formed by its end points and the centre of the circle, and drop a vertical from the centre of the circle to its mid-point. Then, from one of the smaller triangles created,

\(\displaystyle \frac{(a/2)}{r}=\sin(15)\quad \text{so}\quad a=2r\sin(15)\)

where r is the  radius of the circle and the angle (and all following angles) is in degrees.

Similarly,

\(\displaystyle b=2r\sin(45)\quad\text{and}\quad c=2r\sin(\theta),\)

where theta is half of the angle subtended by the chord of length c at the centre of the circle.

Since c = a + b,

\(\displaystyle 2r\sin(\theta)=2r\sin(15)+2r\sin(45) \\\sin(\theta)=\sin(15)+\sin(45)=2\sin(30)\cos(15)\\ \qquad \text{(using the identity sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2) ) }\\ \hspace{25pt} = 2.(1/2).\sin(75) \\ \hspace{25pt}=\sin(75).\)

So the angle subtended by the chord length c at the centre of the circle is 2 times 75 = 150 deg.