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What's the simplest way to draw an triangle (not as a node!) with tikz with a given side and two given angles?

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Interesting, but how about the orientation of triangle in the plane? Should we assume the given edge is parallel to page borders? – Pouya Apr 24 '14 at 13:34
I think you may assume this, because one can easily rotate the result via scope... – student Apr 24 '14 at 13:36
Two or three angles? :) – percusse Apr 24 '14 at 13:47
@percusse, obviously two would suffice. [edit: oops! didn't see the :) face at the end of your comment...] – Pouya Apr 24 '14 at 13:48
up vote 12 down vote accepted

Got an example by googling Solving ASA triangles .

% Given
\draw (0,0) -- (0:\myside)
     let \n1={(180-(\myanga+\myangb))},
         \n2={(\myside*(sin(\myanga)/sin(\n1))} in
     -- (\myangb:\n2) -- cycle node{\n1,\n2};

enter image description here

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Damn you're faster than superman :) – Trefex Apr 24 '14 at 14:08

Not as simple as percusse's answer but easy to understand.

\documentclass[tikz, border=2mm]{standalone}
\usetikzlibrary{intersections, angles, quotes}

\draw coordinate[label=below:a] (a) --++(0:4cm) coordinate[label=below:b] (b);

\path[name path=ac] (a)--++(30:3cm);
\path[name path=bc] (b)--++(180-37:3cm);

\path [name intersections={of = ac and bc, by=c}];

\node[above] at (c) {c};

\draw[use as bounding box] (a)--(b)--(c)--cycle%
     pic[draw, "$30^\circ$", angle eccentricity=1.6] {angle=b--a--c}
    pic[draw, "$37^\circ$", angle eccentricity=1.6] {angle=c--b--a};

enter image description here

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It is too long as a comment but too short for typing exercise.




enter image description here

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