For a long time I've been fascinated by the way you can tile the 80 80 20 isosceles with smaller isosceles/equilateral triangles, like this
I also really like the geometric determination of sin 18 using recursive triangles, which you've epxlored on video.
Have you looked at combining the two? I realised today that the 80 80 20 tiling very quickly and elegantly shows that y=2sin(10) has to satisfy y=1/(3-y^2) ==> y^3-3y^2+1. After adjusting for the factor of 2 this is the same cubic as you talk about on your sin 10 video.
To see this, start with the isosceles tiling from cut-the-knot above, setting the base of the big 80 80 20 triangle to 1; this length then propagates through all the isosceles triangles in the diagram; and set y=2sin(10) as shown.
Because ABC and BCD are similar we know their sides are in proportion so y/1 = 1/AB = 1/(AF+1).
Now focussing on EFA we can add points G and H along FA so that FEH and AEG are congruent to each other and to BCD, so EG = EH =y; and because EGH is similar we know that GH = y^2, so AF = AG +FH - GH = 1+1-y^2 = 2-y^2. Substituting into y=1/(AF+1) gives y=1/(3-y^2) ==> y^3-3y+1=0, QED.
This diagram also gives you the connection with cos40=sin50: the two sides of the big isosceles triangle ABC, namely AB and AC have to be equal. We know AB=3-y^2 as just shown and can see that AC=1+y+2sin50. Equating these gives sin50=1-y/2-(y^2)/2.