Tri-ten-gles!

Let $a,b,c$ be the possible angles divided by $10$ . We are then looking for the number of ordered solution triples $(a,b,c)$ to the equation

$a + b + c = 18$ where $1 \le a \le b \le c.$

(This last inequality statement ensures that we (i) do not count non-degenerate triangles and (ii) do not count the same triangles twice.)

Now look at the separate cases, going from $a = 1$ to $a = 6$ :

• $a = 1$ : we have $8$ solutions; $(1,1,16), (1,2,15), (1,3,14), .... ,(1,8,9).$

• $a = 2$ : we have $7$ solutions; $(2,2,14), (2,3,13), .... ,(2,8,8).$

• $a = 3$ : we have $5$ solutions; $(3,3,12), (3,4,11), ....,(3,7,8).$

• $a = 4$ : we have $4$ solutions; $(4,4,10), (4,5,9), ...., (4,7,7).$

• $a = 5$ : we have $2$ solutions; $(5,5,8), (5,6,7).$

• $a = 6$ : we have $1$ solution; $(6,6,6).$

Adding the tallies up, we have a total of $8 + 7 + 5 + 4 + 2 + 1 = \boxed{27}$ solutions.

We need the number of unordered integer triples (a,b,c) such that $a+b+c=18$ and $a,b,c\ge1$ . If $a=b=c$ we get that $a=6$ so we have N1=1 solution from that.If $a=b\neq c$ we get that $a=\frac{18-c}{2}$ so $c$ needs to be an even number between 2 and 16.So we have N2=8-1=7 solutions from that (within the eight solutions there is also the 6 that we counted in the first case).Now let's count differently to find the number N3 of solutions where each of the integers is different from the other two.With the Stars and Bars method we find that there are $\frac{(18-1)!}{2!(18-1-2)!}=136$ ordered solutions to the original equation.Out of these N1=1 is the one where $a=b=c$ , 3*N2=21 are the ones with two of the numbers being equal (we can arrange (a,a,c) in three different ways) so the other 136-1-21=114 solutions are the ones where the integers are different.We are left with N3=114/6=19 if we ignore the order (we can arrange (a,b,c) in six different ways).So the answer is N=N1+N2+N3=1+7+19=27 solutions.