I think I can completely find those triangles for which $a_{2}=1$.

We have two isosceles triangles forming a third triangle.
Two of their angles must form a 180 degree angle. the two
equal angles of both triangles cannot do this. So the vertex
angle of one triangle must be adjacent to one of the angles
of the second triangle. Let the second triangle have one angle
$A$ and two angles B=$90-A$. Then if the angle of the first
triangle is adjacent to $A$ then It must be of magnitude 180-$A$
and the other two angles of the triangle must be $A/2$ but we then must
add $A/2$ two one of the remaining angles this the triangle with the following
angles:

($A/2$, 90, $90-A/2$) this gives all right triangles which has already been noted in a separate answer.

If the angle of the first triangle is adjacent to $90-A/2$ then we get
a new angle of $45-A/4$ we will have to add this quantity to either
of the two angles giving two triangles with the following angles:

($45-A/4$, A, $135-3A/4$)

($45-A/4$, $90-A/2$, $45+3A/4$)