trying to join to angled plates together,one is straight 140 degrees fold in centre i need to join to sheet plates,one is 1400mm x 265,folded at 140degrees in the middle,the other is 740 by 1400  folded 140 degrees in the middle. This piece needs to slope away from the other at 19 degrees at the fold line....
how do i work out the angles to remove the material so the larger plate slope down to point...
 A: So the folds are parallel to the short sides.  Then I can put the $1400 \times 700$ with the fold flat on the table and each side sloping up at $20$ degrees from the table.  I put one end of the fold in the $1400 \times 265$ touching the end of the fold in the $1400 \times 700$, with the fold also flat on the table, so all of the $1400$ mm folded edges are touching.  Let us call the point where the folds touch $(0,0,0)$ with $+y$ along the $265$ mm fold and $+z$ up.  The $x$ axis is transverse to the folds.
Now, keeping the outer corners touching, I tilt the 265 mm fold up 19 degrees.  The request is to cut a congruent triangle from the end of the fold to all the outer corners to allow the fold ends to touch again.
The corners of the $700$ mm plate are at $(\pm 700 \cos 20,0,700 \sin 20)$.  Rotating the $265$ plate around the origin by $19$ degrees moves its corners to $(\pm 700,-700 \sin 20 \sin 19, 700 \sin 20 \cos 19)$.  If we bisect the segment between the corners we get $(\pm 700, -350 \sin 20 \sin 19,350 \sin 20 (1+\cos 19))$.  Since $(1+\cos 19)/2 \gt 0.97$ we can probably take it to be $1$.  If you rotate the $700$ plate to put one flat plane on the table, this rotates the point to $(\pm 700, -350 \sin 19,0)$.  So cut a triangle off each corner $700 \times 114$ and you should be there.
