I am trying to make a $30$ degree angle by folding an $A4$ paper, but I got nothing. It seems to be easy but I don't have a clue. I can make $A=45,B=67.5,C=67.5$ triangle or make $45 , \frac {45}{2} , \frac{45}{4}, \cdots$ by folding a right angle.
Is there any way to make a triangle with $30,60 ,90 $ degree by folding an $A4$ paper?
Remark: I search Google but there is almost nothing clear.
Remark2: I saw how to make an equilateral triangle, and then fold any angles to half is the answer. But my question is concerned with not using equilateral triangle.

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    $\begingroup$ Why aren't you willing to use an equilateral triangle, exactly? $\endgroup$ Commented Jan 9 at 20:48
  • 1
    $\begingroup$ What's your way to make an equilateral triangle? I can imagine a way to make a 30-60-90 triangle before making a full equilateral triangle. $\endgroup$
    – peterwhy
    Commented Jan 9 at 21:20
  • $\begingroup$ cutoutfoldup.com/125-fold-a-series-of-equilateral-triangles.php this link contains "Fold an Series of Equilateral Triangles" $\endgroup$
    – Khosrotash
    Commented Jan 10 at 19:17

2 Answers 2

  1. Mark the centre line parallel to the long sides, by folding;
  2. Fold one corner to this centre line, so that short side has one end on the centre line and one end in the original position;
  3. This crease forms the hypothenuse of a 30-60-90 triangle.

The aspect ratio of the paper does not matter.

Back-of-the-envelope demo, using a similar C5 envelope:

Back-of-the-envelope demo, using a similar C5 envelope

  • $\begingroup$ Not trivial, you need to mark the triangle, that has its half side, and hypotenuse of 1. However, it seems the method works for any size of sheet, not A-series only. $\endgroup$
    – dEmigOd
    Commented Jan 10 at 6:34
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    $\begingroup$ Best use of the idiom "back of the envelope" :) $\endgroup$
    – justhalf
    Commented Jan 10 at 7:54
  • $\begingroup$ @dEmigOd At the lower right corner: After the second fold, the (moved) short side and the long side make $30^\circ$. (This answers the first part of the question.) The second crease also bisects the complementary $60^\circ$ angle, makes $30^\circ$ with the short side, and forms a 30-60-90 triangle with the paper sides. $\endgroup$
    – peterwhy
    Commented Jan 10 at 23:04

A $30^\circ$-$60^\circ$-$90^\circ$ triangle has its short leg half the length of its hypotenuse. So, you just need to have a sidelength of length $1/2$ and a hypotenuse of length $1$. First make a square, then fold it in half to get a rectangle. Then, fold one of the corners so it touches the opposite side.

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