Such triangles exist. I found one 3:8:10, where the angle opposite 8 is 1/3 of that opposite 10. The trick here is to pick really small primes in the form of $x+y\sqrt{-n}$, and cube the result. Note here that we're using $n=7$, and the prime is $1\frac 12 + \frac 12\sqrt{-7}$. This is a pretty tiny cube, one gets then a matrix
( 3 -7 ) (3) (2) (36)
( 1 3 ) (1) (6) (20)
A B = A^3
You then divide through by common factors, to get coordinates at $A =0,0$ $B=+6,0$, and $C=-9,5\sqrt{7}$. The three sides are AB=6, AC=16, and BC=20, which gives the indicated triangle.
HINT: Let $a,b,c$ denote the lengths of the sides of the triangle. Then we should have
$$a+b>c\quad\mbox{and}\quad a+b+c=n.$$
The above two equations implies $a+b-c=n-2c>0$, i.e. $c<n/2$ and similarly $a,b<n/2$. So you need to count the number of triples $(a,b,c)$ such that $0<a\leq b\leq c$ and $a,b,c<n/2$.
EDIT: Note that you only need to find the number of tuples $(a,b)$ such that $0<a\leq b$ and $a,b<n/2$, because $c$ is fixed once you specify $a$ and $b$. Let us consider two cases.
Case (i) $n$ is even: For a fixed $a$, you have the following choices for $b$: $a+1, a+2,\cdots ,n/2-1$, that is $n/2-1-a$ choices. Now $a$ can take values $1,2,\ldots ,n/2-1$. So the number of triangles in this case is
$$\sum_{a=1}^{n/2-1}(n/2-1-a)=(n/2-2)+(n/2-3)+\cdots +1=\frac{(n/2-2)(n/2-1)}{2}.$$
Case (ii) $n$ is odd: For a fixed $a$, you have the following choices for $b$: $a+1, a+2,\cdots ,(n-1)/2$, that is $(n-1)/2-a$ choices. Now $a$ can take values $1,2,\ldots ,(n-1)/2$. So the number of triangles in this case is
$$\sum_{a=1}^{(n-1)/2}((n-1)/2-a)=((n-1)/2-1)+\cdots +1=\frac{((n-1)/2-1)(n-1)/2}{2}.$$
Best Answer
The number of triangles with perimeter $n$ and integer side lengths is given by Alcuin's sequence $T(n)$. The generating function for $T(n)$ is $\dfrac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$. Alcuin's sequence can be expressed as
$$T(n)=\begin{cases}\left[\frac{n^2}{48}\right]&n\text{ even}\\\left[\frac{(n+3)^2}{48}\right]&n\text{ odd}\end{cases}$$
where $[x]$ is the nearest integer function, and thus $T(36)=27$. See this article by Krier and Manvel for more details. See also Andrews, Jordan/Walch/Wisner, these two by Hirschhorn, and Bindner/Erickson.