Question 1035358
Perimeter of the triangle:  {{{35.3+36.9+56.9=35.3+36.9+L+r}}}
Along the ground:  {{{L+r=56.9}}}


Triangle Area:  {{{(1/2)*(56.9)=(1/2)(L+r)}}}, or {{{56.9=L+r}}}


Only one equation for that much, being {{{highlight_green(L+r=56.9)}}}.


The left and right side triangles each are Right triangles with height h.
{{{system(h^2+L^2=35.3^2,h^2+r^2=36.9^2)}}}


Equating h^2:
{{{35.3^2-L^2=36.9^2-r^2}}}
{{{35.3^2-L^2+r^2=36.9^2}}}
{{{r^2-L^2=36.9^2-35.3^2}}}
{{{r^2-L^2=1361.61-1246.09}}}
{{{r^2-L^2=115.52}}}


Summarizing as a system of equations
{{{system(L+r=56.9,r^2-L^2=115.52)}}}
-
What to do next is solve this system for L and r;
and then use simple trigonometry to find the angles asked.  Cosines would be best.