# Simple way to draw a triangle with one side and two angles given

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

Got an example by googling Solving ASA triangles .

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
% Given
\def\myside{9}
\def\myanga{76}
\def\myangb{34}
%===============
\draw (0,0) -- (0:\myside)
let \n1={(180-(\myanga+\myangb))},
\n2={(\myside*(sin(\myanga)/sin(\n1))} in
-- (\myangb:\n2) -- cycle node{\n1,\n2};
\end{tikzpicture}
\end{document}


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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}

\begin{document}
\begin{tikzpicture}
\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};
\end{tikzpicture}
\end{document}


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

\documentclass[pstricks,border=24pt]{standalone}
\usepackage{pst-eucl}

\newpsstyle{nolabel}{PointName=none,PointSymbol=none}

\begin{document}
\begin{pspicture}(8,3)
\pstGeonode[PosAngle=-90]{A}(8,0){B}
\pstRotation[style=nolabel,RotAngle=30]{A}{B}
\pstRotation[style=nolabel,RotAngle=-37]{B}{A}
\pstInterLL[PosAngle=90]{A}{B'}{B}{A'}{C}
\pspolygon(A)(B)(C)
\pstMarkAngle{B}{A}{C}{$30^\circ$}
\pstMarkAngle{C}{B}{A}{$37^\circ$}