Consider the sum of the angles of the triangles at each type of the vertices. At type 0, the sum is $180^\circ$. At type 1, the sum is $180t_1$. At type 2, the sum is $360t_2$. Therefore
$$180 + 180t_1 + 360t_2 = 180n \Rightarrow 1 + t_1 + 2t_2 = n.$$
Now each Type 1 vertex has at most 1 angle of $x^\circ$ and each Type 2 vertex has at most 2 which each Type 0 vertex has, none. So the number of angles of $x^\circ$ is $\le t_1 + 2t_2 < n$, a contradiction.

Now we construct for each $x \in (0, 120]$. It is trivial for $x = 120$. Take the centre of the triangle and join it to the vertices to obtain a partition into 3 congruent triangles. Now assume $x \in (0, 120)$.

Define an $(a, b)$-trapezium to be the trapezium $ABCD$ with $\angle A = \angle D = 60^\circ$, $AB = CD = a$, $BC = b$. Then it follows that $AD = a + b$. We call a polygon $x$-good if it can be partitioned into triangles each with one $x^\circ$ angle.

Lemma 1: There exists $R(a)$ such that an $(a, b)$-trapezium is $x$-good when $b > R(a)$.
Proof: The proof is clear from Fig. A. The second, third and fourth triangles are isosceles. The fifth and sixth triangles form a parallelogram with arbitrarily long horizontal side.

Fig.A

Fig.B

Lemma 2: An $(a, b)$-trapezium is $x$-good.
Proof: It follows from the fact that the trapezium can be sliced into arbitrarily thin $(a', b')$-trapezium so that $b' > R(a')$. (See Fig. B.)
Since an equilateral triangle can be partitioned in $(a, b)$-trapeziums, the proof is complete.