I produced a diagram much like the one you sent and then I put some labels on it.
The initial triangle has base of length b and height h so the area is given by
A = 1/2 b h
The horizontal line cuts off a certain proportion of the area above the line. I want to call this proportion p. (I would rather work with proportions than percentages.) The horizontal cut-off line is of length b' and the triangle above this line has height h'. Thus the area of the triangle above the cut-off line is given by
pA = 1/2 b' h'
Substitute A from the equation above to get
p 1/2 b h = 1/2 b' h'
p b h = b' h' or
p = b'/b h'/h
The original triangle and the triangle above the cut-off line are similar so
b'/b = h'/h
p = h'/h h'/h = (h'/h)2
h' = √p h
Thus in your example you want 15% of the area above the cut-off line so p = 0.15 and √0.15 = 0.39. Thus draw the cut-off line so that the height of the triangle above the cut-off line is 39% of the height of the original triangle.
The next cut-off line is to leave 40% of the area of the original triangle above it so here p = 0.40 and √0.40 = 0.63. Thus draw the next cut-off line so that the height of the triangle above this cut-off line is 63% of the height of the original triangle.