if the triangle is ABC, then side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.
you can use the law of cosines to find angle C.
the law of cosines is c^2 = a^2 + b^2 - 2abcos(C).
that becomes 13^2 = 8^2 + 11^2 - 2*8*11*cos(C).
add 2*8*11*cos(C) to both sides of the and subtract 13^2 from both sides of the equation to get 2*8*11*cos(C) = 8^2 + 11^2 - 13^2.
divide both sides of the equation by (2*8*11) to get cos(C) = (8^2 + 11^2 - 13^2) / (2 * 8 * 11) = .0909091909.
arcos(.0909090909) = 84.78409143 degrees
that's angle C.
you can use the law of sines to find angle B.
that law says sin(C) / c = sin(B) / b.
that becomes sin(84.78409143)/13 = sin(B)/11.
solve for sin(B) to get sin(B) = sin(84.78409143)/13 * 11 = .8426500885.
arcsin(.842650885) = 57.42102961 degrees.
since the sum of the interior angles of a triangle = 180 degrees, then angle A = 180 - 84.78409143 - 57.42102961 = 37.79487896.
the largest angle is opposite the largest side and the smallest angle is opposite the smallest side with the middle angle opposite the middle side.
here are the results in a table of values.
side sine of angle angle 8 .612836428 37.79487896 11 .8426500885 57.42102961 13 .9958591955 84.78409143
sum of the angles is 180 as it should be.
here's my diagram
