Arectangle has dimensions 10 cm by 6 cm. determine the measures of the angles at the point of where the diagonals intersect

See the attached figure to better understand the problem

we know that

in the triangle ABC
tan(alfa/2)=AC/BC----> 5/3
so
(alfa/2)=arctan (5/3)---------> 59.0362°
alfa=2*59.0362--------> alfa=118.07°
alfa+beta=180°----------> because are supplementary angles
beta=180-118.07-------> beta=61.93°

the answer is 
the angles are
alfa=118.07°
beta=61.93°