Thanks Gino,
You have done a lot of work there but unfortunately, you did label O incorrectly.
\(15^2=m^2+t^2-2mt\cdot cos\alpha\\ Cos\alpha = \frac{2t}{4m}=\frac{t}{2m}\\ so\\ 225=m^2+t^2-2mt \cdot \frac{t}{2m}\\ 225=m^2\\ m=15 \)
This means that the hypotenuse is 60
and the sum of the other 2 sides is 112-60 =52
As guest has already told us very loudly , this is impossible.
So such a triangle cannot exist.