My question is: " Find the total number of non similar triangles which can be formed such that all the angles of the triangles are integers"
My attempt: Let $x$, $y$ and $z$ be the angles of the triangles. So $x+y+z=180$, where $x,y,z \ge1$. Total =$\binom{n-1}{r-1}=\binom{179}{2}=15931$. How to proceed further? We have to subtract the total number of similar triangles from this number.
I'm assuming you measure angles in degrees, that degenerate triangles are forbidden, and that mirror congruent triangles are considered congruent.
The similarity type of a triangle is then determined by its angles $x$, $y$, $z$, whereby $x\leq y\leq z$. It follows that $1\leq x\leq 60$. Given $x$ we can write $y=x+t$ with $t\geq0$, and we have to take care that $$x+t=y\leq z=180-x-(x+t)\ .$$ It follows that $$0\leq t\leq {180-3x\over2}\ ,$$ so that we have $$a(x):=1+\left\lfloor{180-3x\over2}\right\rfloor$$ admissible choices for $t$. For $x=2m$ we obtain $a(2m)=91-3m$, and for $x=2m-1$ we obtain $a(2m-1)=92-3m$. It follows that the total number $N$ of triangles is given by $$N=\sum_{m=1}^{30}(183-6m)=2700\ .$$