What is a simple way to locate the foci of an ellipse? I have DIY project going, and ran into the same issue as posted in the question here:  stove jack for a tent
The answer was helpful, but somewhere along the line I got lost in the math... (Perhaps because the example was over simplified, using a 45 triangle with units of 1, and the distance between foci equaled the minor axis?)
Because my roof pitch isn't 45 degrees, I wanted to work the problem on a 30-60-90 triangle to make sure I understood the concept before I unfold my tent and take measurements of the actual slope.  I plan on using a 4" diameter stove pipe, and created the diagram below.  (apologies for the image quality)

I realize that I could make a template using a French curve like I did for my diagram above and be "close enough", but I would like to try the string method as described in the answer on the other question I linked to.
Is there a relatively easy way to calculate distance X given the angles, and/or dimensions for someone like me who is not a math whiz?
ADDENDUM:
I am accepting Parcly Taxel’s answer because it was concise, first, correct, and it pointed me in the right direction even if it didn’t state the obvious.
I understand that equations are the language of mathematics, and I didn’t ask for a plain English answer, but as I alluded to in a comment – for many of us knowledge of anything beyond the basic arithmetic we use regularly may have been hard won, and could have atrophied in the intervening decades.  Even equations that are very simple for the left brained intelligentsia frequenting this venue may create anxiety for us simpletons cutting holes in our tents who need to review whether to use COS or TAN to calculate the opposite side!
I don’t want this to morph into a critique about how I was taught math, but now that I “get it” I wanted to offer an example below of how I might answer the question for an admitted non-math person.  FWIW, and I welcome any feedback in the comments…

Using standard symbols of an ellipse:
For A=$30^0$
$$a= b \sec A\quad = b \cdot \dfrac{2}{\sqrt3}; \quad c^2=a^2-b^2 =\dfrac{b^2}{3}$$
Take square root and double it for inter-focal distance
$$ c=\dfrac{b}{\sqrt3},\quad 2c =\dfrac{2b}{\sqrt3} \approx 2.31\text{  for b=}2^{''.}$$
