I already submitted this solution but I later realised I had skipped an important conclusion so I have just added it here.
If we name angle BAC angle x, we can say that angle ABC is equal to 180-2x as base angles in an isosceles triangle are equal and angles in a triangle add up to 180 degrees. Then angle DBC must equal 2x as the exterior angle is equal to the sum of the two interior angles. So angle BCD must equal 180-4x because base angles in an isosceles triangle are equal and angles in a triangle add up to 180 degrees. Then angle DCE is equal to 3x as angles BCA+BCD+DCE add up to 180 as angles on a straight line sum to 180 degrees and we know BCA and BCD so if we subtract them then we are left with DCE=3x. then CDE=180-6x for the same reason that BCD=180-4x and FDE=4x for the same reason that DCE=3x. Here we can see that there is a repeating pattern that the base angles in each isosceles triangle going upwards is increase by x each time. If we follow this pattern up to the top of the 'main' isosceles triangle then we get that angle GIH=7x. Since triangle AHI is isosceles then AHI must also be 7x as base angles in an isosceles triangle are equal. As angles in a triangle sum to 180 degrees then we get 15x=180 so x=12 degrees. So angle HAI is 12 degrees and angles AHI and AIH are both 84 degrees. For the extensions, the base angles for the isosceles triangle will be nx so the sum of all the angles in the isosceles triangle will be 2nx+x=180. We know that the base angles will be nx as there are n-1 triangles other than BAC so the pattern occurs that many times. Each time the angle increases by x so the angle you end up with would be (n-1)x+x which equals nx. So x in terms of n would be x=180/(2n+1). So as long as 180 is divisible by 2n+1 then x will be a whole number and therefore all the angles will be a whole number as n will be a whole number and their product will be whole as well. An example of a possible value of n which fulfills the conditions would be n=4.