Question 980796
While I was busy drawing pretty pictures, Tim was posting a more succinct answer.
I am posting my answer too, hoping that my drawings, and my different angle (pun intended) help too.
{{{drawing(300,100,-45,135,-15,45,
triangle(0,0,-37.65,35.865,93.35,35.865),
triangle(131,0,0,0,93.35,35.865),
green(arrow(0,0,131,0)),green(arrow(0,0,-37.65,35.865)),
red(arrow(0,0,93.35,35.865)),locate(46.7,17.9,red(100)),
locate(65,0,green(131)),locate(29,35.8,131),
locate(-27,20,green(52)),locate(103,22,52),
green(arc(0,0,40,40,-136.4,0)),locate(0,19,green(theta)),
arc(131,0,40,40,-180,-136.4),locate(114,10,A)
)}}} I am assuming that you have to find the angle {{{green(theta)}}} ,
but we can also find the other angle in the parallelogram,
which is supplementary to {{{green(theta)}}} : {{{A=pi-green(theta)}}} .
The resultant divides that parallelogram into two (congruent) triangles.
We need to apply to one of those triangles the law of cosines,
which is sort-of like an extension of the Pythagorean theorem,
for triangles that are not necessarily right triangles.
For {{{drawing(300,100,-45,135,-15,45,
triangle(131,0,0,0,93.35,35.865),
green(line(0,0,131,0)),
red(line(0,0,93.35,35.865)),locate(46.7,20,red(a)),
locate(65,0,green(b)),
locate(109,28,c),
arc(131,0,40,40,-180,-136.4),locate(115,10,A)
)}}} law of cosines says {{{a^2=b^2+c^2-2*b*c*cos(A)}}} .
Applying law of cosines to the triangle above, with {{{system(a=100,b=131,c=52,A=pi-green(theta))}}} , we get
{{{100^2=131^2+52^2-2*131*52*cos(A)}}} and form that we get
{{{cos(A)=(131^2+52^2-100^2)/(2*131*52)=0.724090}}} (rounded)
Since they are supplementary angles {{{cos(green(theta))=-cos(A)=-0.724090}}} (rounded) .
Inverse cosine function tells us that the angles, in radians are
{{{green(theta)=2.3805}}} and {{{A=0.7611}}}