Best way to denote an angle in TikZ?

Question

What is the best and simplest way to denote an angle, with an arc and a label telling its name? I wish to be able to control (obviously) the vertices on which the angle lays, the name of the angle and also control the radius of the little arc and whether it should fill the whole part-of-a-circle shape in color.

Answer 1

I just have a look to this question and related ones but I don't see anywhere my solution to mark an angle when you don't know its measure (maybe I haven't read all the post about the topic...).

So here it is:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}
\coordinate[label=below left:$A$] (A) at (0,0);
\coordinate[label=below right:$x$] (X) at (6,1);
\coordinate[label=above left:$y$] (Y) at (3,5);

\draw[thick] (X) -- (A) -- (Y);

% Mark the angle XAY
\begin{scope}
\path[clip] (A) -- (X) -- (Y);
\fill[red, opacity=0.5, draw=black] (A) circle (5mm);
\node at ($(A)+(30:7mm)$) {$\theta$};
\end{scope}

\end{tikzpicture}

\end{document}

I draw a full circle and I hide the bad part of it by clipping with a path built with the three points which define the angle. It seems so easy that it might be a bad way to do it but it works pretty well. Here I put manually the angle measure to place \theta but it could be automated surely.

Please, feel free to comment it if you think there are some situations where it doesn't work!

Answer 2

Another example with TikZ only . The line important is \draw [color=black](a)+(10:1.2) node[rotate=0] {$\beta$}; the label for the angle is placed with the help of the center and this center is translated (a)+(10:1.2) .

\documentclass[11pt]{scrartcl}
\usepackage{tikz}
\usetikzlibrary{through,calc}

\begin{document}
\begin{tikzpicture}[scale=.8]
\coordinate (o) at (3,0);
\coordinate [label=150:$A$](a) at (0,0);
\coordinate [label=0:$B$](b) at (6,0);
\coordinate[label=90:$C$] (c) at (4,5);
\draw (a) -- (b) -- (c) -- cycle;
\node(c1) at (o)[draw,circle through=(a)] {};
\coordinate[label=90:$D$] (d) at (intersection 1 of c1 and a--c);
\coordinate[label=80:$E$] (e) at (intersection 1 of c1 and b--c);
\coordinate[label=60:$F$] (f) at (intersection  of a--e and b--d);
\coordinate[label=-90:$H$] (h) at ($(a)!(c)!(b)$);
\fill[red] let \p1=(f),\p2=(h) in (\x2,\y1) circle (2pt);
\draw[blue] (c) -- ($(a)!(c)!(b)$);
\draw [blue,very thick](b) +(142:.8cm) arc (142:180:.8cm);
\draw [color=black](b)+(160:1) node[rotate=0] {$\alpha$};
\draw [green!50!black,very thick](a) +(0:.8cm) arc (0:21:.8cm);
\draw [color=black](a)+(10:1.2) node[rotate=0] {$\beta$};
\draw [red!50!black,very thick](a) +(21:.8cm) arc (21:53:.8cm);
\draw [color=black](a)+(32:1.1) node[rotate=0] {$\gamma$};
\draw [red!50!black,very thick](b) +(110:.8cm) arc (110:142:.8cm);
\draw [color=black](b)+(128:1) node[rotate=0] {$\gamma$};
\draw [orange,very thick](c) +(-70:.8cm) arc (-70:-127:.8cm);
\draw [color=black](c)+(-100:1.2) node[rotate=0] {$\delta$};
\draw [orange,very thick](f) +(143:.8cm) arc (143:203:.8cm);
\draw [orange,very thick](f) +(24:.6cm) arc (24:-38:.6cm);
\draw (a) -- (e);
\draw (d) -- (b);
\draw[anchor=base,color=blue] (h.center)  ++(.3,0)  -- ++(0,0.3) -- ++(-0.3,0);
\end{tikzpicture}
\end{document}

Answer 3

I took the PolGab's code and changed it a little:

1. Put an optional argument to change the color
2. Change order of points to name the angle (usually in angle ABC, B is the origin...)
3. Automate the position of the label.

Here is the code:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\newcommand\markangle[6][red]{% [color] {X} {origin} {Y} {mark} {radius}
  % filled circle: red by default
  \begin{scope}
    \path[clip] (#2) -- (#3) -- (#4);
    \fill[color=#1,fill opacity=0.5,draw=#1,name path global=circle]
    (#3) circle (#6mm);
  \end{scope}
  % middle calculation
  \path[name path=line one] (#3) -- (#2);
  \path[name path=line two] (#3) -- (#4);
  \path[%
  name intersections={of=line one and circle, by={inter one}},
  name intersections={of=line two and circle, by={inter two}}
  ] (inter one) -- (inter two) coordinate[pos=.5] (middle);
  % bissectrice definition
  \path[%
  name path=bissectrice
  ] (#3) -- (barycentric cs:#3=-1,middle=1.2);
  % put mark
  \path[
  name intersections={of=bissectrice and circle, by={middleArc}}
  ] (#3) -- (middleArc) node[pos=1.3] {#5};
  }
\begin{document}
\begin{tikzpicture}
\coordinate[label=below left:$A$] (A) at (0,0);
\coordinate[label=below right:$B$] (B) at (2,0);
\coordinate[label=above:$C$] (C) at (5,5);
\draw[thick] (B) -- (A) -- (C) -- cycle;
\markangle{A}{B}{C}{$\beta$}{5}
\markangle[blue]{B}{A}{C}{$\alpha$}{6}
\markangle[green]{B}{C}{A}{$\gamma$}{12}
\end{tikzpicture}
\end{document}

It will be great if we could pass the radius of the mark as an optional argument also (for instance 5mm by default). Thus, we could use either \markangle{A}{B}{C} if we want to use default values for color and mark radius or \markangle[blue, 10]{A}{B}{C} if we don't. But I think it is impossible because a \newcommand just takes one optional argument, right?

Answer 4

(Edit: recent version of Tikz requires name path global instead of name path.)

Here is a complement to sylcha's answer.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\newcommand\markangle[6]{% origin X Y radius radiusmark mark
  % fill red circle
  \begin{scope}
    \path[clip] (#1) -- (#2) -- (#3);
    \fill[color=red,fill opacity=0.5,draw=red,name path global=circle]
    (#1) circle (#4);
  \end{scope}
  % middle calculation
  \path[name path=line one] (#1) -- (#2);
  \path[name path=line two] (#1) -- (#3);
  \path[%
  name intersections={of=line one and circle, by={inter one}},
  name intersections={of=line two and circle, by={inter two}}
  ] (inter one) -- (inter two) coordinate[pos=.5] (middle);
  % put mark
  \node at ($(#1)!#5!(middle)$) {#6};
}
\begin{document}
\begin{tikzpicture}
\coordinate[label=below left:$A$] (A) at (0,0);
\coordinate[label=below right:$x$] (X) at (2,0);
\coordinate[label=above left:$y$] (Y) at (-3,5);
\draw[thick] (X) -- (A) -- (Y);
\markangle{A}{X}{Y}{7mm}{5mm}{$\theta$}
\end{tikzpicture}
\end{document}

Answer 5

I have developped 2 macros that I think can help you:

\newcommand{\Angulo}[7]{
\draw[#1]
    let \p1=($(#3) - (#4)$), \p2=($(#5) - (#4)$),
    \n{aIni}={atan2(\x1,\y1)},
    \n{aFin}={atan2(\x2,\y2)},
    \n{aMed}={0.5*\n{aIni}+0.5*\n{aFin}}
    in ($(#4)!#2 * \longU!(#3)$) arc (\n{aIni}:\n{aFin}:#2 * \longU)
    node at ($(#4) + (\n{aMed}:#2 * \longU)$) [#6] {#7};
}

\newcommand{\AnguloC}[7]{
\draw[#1]
    let \p1=($(#3) - (#4)$), \p2=($(#5) - (#4)$),
    \n{aIni}={atan2(\x1,\y1)},
    \n{aFin}={360+atan2(\x2,\y2)},
    \n{aMed}={0.5*\n{aIni}+0.5*\n{aFin}}
    in ($(#4)!#2 * \longU!(#3)$) arc (\n{aIni}:\n{aFin}:#2 * \longU)
    node at ($(#4) + (\n{aMed}:#2 * \longU)$) [#6] {#7};
}

\begin{tikzpicture}
%      Parameters:
%      \Angulo{drawing style}{radius}{starting point}{center point}{ending point}{text style}{text};
%      You can use coordinates or node names as points.

  \coordinate (nodeCenter) at (0,0);
  \Angulo{blue,->}{1}{10,0}{nodeCenter}{0,10}{above right}{$\alpha$};
  \AnguloC{red,<-}{1}{0,10}{nodeCenter}{10,0}{above right}{$\alpha - 360$};
\end{tikzpicture}