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Werner_E
Werner_EAnswer
25-Diamond I
February 15, 2021

OK, so the goal is not (or not only) drawing, but rather ....??

Do you need a function of some kind which also gives you any point on the sides of the triangle, not just the vertices? Or is Luc's approach already doing what you have in mind?

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9 replies

LucMeekes23-Emerald IV
23-Emerald IV
February 15, 2021

LucMeekes_0-1613404002701.png

 

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Luc

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-MFra-21-Topaz IIAuthor
21-Topaz II
February 15, 2021

Oh! yes nice,

but my version considers the height of the triangle, changing which, for example, I get an isosceles triangle. however, the use of other geometric figures to build the triangle is not welcome.

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LucMeekes23-Emerald IV
23-Emerald IV
February 15, 2021

The general case, for an n-angle and orientation o, is:

LucMeekes_1-1613406473939.png

 

 

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Luc

LM_20210215_Triangle.gif

 

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    Werner_E
    Werner_E25-Diamond I
    25-Diamond I
    February 15, 2021

    Not sure what exactly you are looking for.

    If your goal is to just draw an equilateral triangle dependent on the side length and in the same position in the coordinate system as yours, a simple way would be this:

    Werner_E_0-1613407033414.png

    As Luc wrote, three points are all you need - actually four as you have to duplicate the first one to get a closed figure. You can leave it to Mathcad's 2D plot to connect the points.

     

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    Werner_E
    Werner_E25-Diamond I
    25-Diamond I
    February 15, 2021

    Just read your answer to Luc's post where you say that you also want to draw isosceles triangles depending on the base length and the height.

    Here it is:

    Werner_E_0-1613407655339.png

     

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