數學問題......10 points

let the points of the parallelogram be ABCD respectively
cut the parallelogram into two triangle
each triangle's area is 80/2 = 40c㎡
ABC is an acute angle
area of triangle = 1/2 (15) (6) (sin ABC)
40 = 45 sin ABC
ABC = 62.7 °
the another angle = 180-62.7=117.3°
the angles of the parallelogram are 62.7° and 117.3°

Area of a parallelogram = ab sinx
ab sinx=6x15 sinx=80
90 sinx=80
sin x=8/9
x=sin^-1 8/9
x=62.7
Therefore,
the angles of the parallelogram are 62.7 and 117.3.