Question 574123
{{{drawing(300,100,-7,14,-1.5,5.5,
triangle(0,0,13,0,-6.423,4.769), locate(-0.5,0,A),
locate(5,0,13m), locate(-4,2.2,8m),locate(3.3,3.5,20m),
green(line(-6.423,4.769,6.577,4.769)),green(line(13,0,6.577,4.769))
)}}} There is your picture. I completed the parallelogram in green, so you can see it, but you are concerned with just the half of the parallelogram that I drew as a black triangle.
Since you know only the length of all three sides of the triangle, you are forced to use law of cosines to solve it. Applied to this triangle, law of cosines says that:
{{{8^2+13^2-2*8*13*cos(A)=20^2}}} --> {{{64+169-208cos(A)=400}}} --> {{{cos(A)=(400-64-169)/(-208)}}} --> {{{cos(A)=-167/208}}} --> {{{A=2.503}}} radians or {{{A=143.4^o}}}
In a parallelogram, the opposite angle (green in this case) is congruent, so it has the same measure.
The other two angles are supplementary, so their measure is 
{{{180^o-143.4^o=36.6^o}}}