The two arms are holding the box along its mid at points E(2.5,12) and F(7.5,12).
Both the triangles(ACE & BDF) shown are isosceles.
Draw a perpendicular from vertex C(D) to the opposite side AE(BF) of the triangles which would also be the line bisector of the side AE(BF).
Calculate the cosine inverse of the ratios of the sides and also the side lengths:-
AC = CE = BD = DF = 15;
AE = 12.26;
beta = arccos(An/CA);
angle(EAG) = arctan(EG/AG);
angle(FBH) = arctan(HF/BH);
alpha = arccos(BM/BD);

Thus calculated angles are:-
theta1 = 144.11*
theta2 = -131.76*
theta3 = 81.45*
theta4 = 109.43*