Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$ Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.