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I am able to draw a nice right triangle with
\begin{tikzpicture}[thick]
\draw (0,0) -- (4,0) node[midway,below]{$f$} -- (4,2) node[midway,right]{$\frac{x}{2}$}-- cycle;
\end{tikzpicture}
Now I want to label the left angle of the triangle and I am stuck on how to do this.
My answer shows how to label the vertex and the angle. The macro tikzAngleOfLineis not optimized
\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\newcommand{\tikzAngleOfLine}{\tikz@AngleOfLine}
\def\tikz@AngleOfLine(#1)(#2)#3{%
\pgfmathanglebetweenpoints{%
\pgfpointanchor{#1}{center}}{%
\pgfpointanchor{#2}{center}}
\pgfmathsetmacro{#3}{\pgfmathresult}%
}
\begin{tikzpicture}
\coordinate (A) at (1,1);
\coordinate (B) at ($(A)+(25:3)$);
\coordinate (C) at ($(A)+(100:5)$);
\draw (A) node[left]{$A$} -- (B) node[right]{$B$}node[midway,below]{$c$} -- (C)node[above]{$C$}node[midway,above]{$a$} -- (A)node[midway,left]{$b$};
\tikzAngleOfLine(A)(B){\AngleStart}
\tikzAngleOfLine(A)(C){\AngleEnd}
\draw[red,<->] (A)+(\AngleStart:2cm) arc (\AngleStart:\AngleEnd:2 cm);
\node[circle,fill=green] at ($(A)+({(\AngleStart+\AngleEnd)/2}:1 cm)$) {$\alpha$};
\end{tikzpicture}
\end{document}
Here is a solution using TikZ 3.0:
\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{quotes,angles}
\begin{document}
\begin{tikzpicture}
\draw
(3,-1) coordinate (a) node[right] {a}
-- (0,0) coordinate (b) node[left] {b}
-- (2,2) coordinate (c) node[above right] {c}
pic["$\alpha$",draw=orange,<->,angle eccentricity=1.2,angle radius=1cm] {angle=a--b--c};
\end{tikzpicture}
\end{document}
\usetikzlibrary{babel} in the preamble (See tex.stackexchange.com/questions/166772/…).
Jul 22, 2014 at 11:57
I've built on the excellent answer from Alain Matthes above and defined a more direct macro. I've also included an example of its use case.
\newcommand{\tikzAngleOfLine}{\tikz@AngleOfLine}
\def\tikz@AngleOfLine(#1)(#2)#3{%
\pgfmathanglebetweenpoints{%
\pgfpointanchor{#1}{center}}{%
\pgfpointanchor{#2}{center}}
\pgfmathsetmacro{#3}{\pgfmathresult}%
}
\newcommand{\tikzMarkAngle}[3]{
\tikzAngleOfLine#1#2{\AngleStart}
\tikzAngleOfLine#1#3{\AngleEnd}
\draw #1+(\AngleStart:0.15cm) arc (\AngleStart:\AngleEnd:0.15cm);
}
\begin{tikzpicture}[scale=4,line width=1pt]
\coordinate (B) at (0,0);
\coordinate (C) at (1,0);
\coordinate (A) at (0.5,1.5);
\draw (A) -- (B) -- (C) -- cycle;
\node [below left] at (B) {B};
\node [below right] at (C) {C};
\node [above] at (A) {A};
\coordinate (D) at (barycentric cs:A=0.5 ,C=0.5);
\node [above right] at (D) {D};
\draw (B) -- (D);
\tikzAngleOfLine(A)(C){\AngleStart}
\tikzAngleOfLine(A)(B){\AngleEnd}
\draw (A)+(\AngleStart:0.15cm) arc (\AngleStart:\AngleEnd:0.15cm);
\tikzMarkAngle{(C)}{(B)}{(A)}
\tikzMarkAngle{(D)}{(A)}{(B)}
\end{tikzpicture}
Here's an example use of the macro.
I've also included a slightly more involved example below.
\newcommand{\tikzAngleOfLine}{\tikz@AngleOfLine}
\def\tikz@AngleOfLine(#1)(#2)#3{%
\pgfmathanglebetweenpoints{%
\pgfpointanchor{#1}{center}}{%
\pgfpointanchor{#2}{center}}
\pgfmathsetmacro{#3}{\pgfmathresult}%
}
\newcommand{\tikzMarkAngle}[4][0.15cm]{
\tikzAngleOfLine#2#3{\AngleStart}
\tikzAngleOfLine#2#4{\AngleEnd}
\draw[<->] #2+(\AngleStart:#1) arc (\AngleStart:\AngleEnd:#1);
}
\begin{tikzpicture}[scale=4,line width=1pt]
\coordinate (B) at (0,0);
\coordinate (C) at (1,0);
\coordinate (A) at (0.5,1.5);
\draw (A) -- (B) -- (C) -- cycle;
\node [below left] at (B) {B};
\node [below right] at (C) {C};
\node [above] at (A) {A};
\coordinate (D) at (barycentric cs:A=0.5 ,C=0.5);
\coordinate (E) at (barycentric cs:A=0.40 ,B=0.60);
\node [above right] at (D) {D};
\node [above left] at (E) {E};
\draw (B) -- (D);
\draw (C) -- (E);
\draw (E) -- (D);
\tikzMarkAngle{(A)}{(B)}{(C)}
\node at ($(A)+(-90:0.23)$) {$20^\circ$};
\tikzMarkAngle{(B)}{(D)}{(C)}
\node at ($(B)+(25:0.23)$) {$60^\circ$};
\tikzMarkAngle{(C)}{(B)}{(E)}
\node at ($(C)+(155:0.23)$) {$50^\circ$};
\begin{scope}[color=orange]
\tikzMarkAngle{(D)}{(E)}{(B)}
\node at ($(D)+(-20:0.15)$) {?=$30^\circ$};
\end{scope}
\tikzMarkAngle{(B)}{(A)}{(D)}
\node at ($(B)+(55:0.23)$) {$20^\circ$};
\tikzMarkAngle{(C)}{(E)}{(D)}
\node at ($(C)+(120:0.23)$) {$30^\circ$};
\coordinate (F) at (barycentric cs:A=0.50 ,B=0.50);
\node [above left] at (F) {F};
\draw (F) -- (D);
\draw (C) -- (F);
\coordinate (G) at (intersection of D--B and C--F);
\node [above] at (G) {G};
\draw (E) -- (G);
\begin{scope}[color=blue]
\tikzMarkAngle{(G)}{(F)}{(E)}
\node at ($(G)+(145:0.20)$) {$40^\circ$};
\tikzMarkAngle[0.07cm]{(F)}{(E)}{(G)}
\node at ($(F)+(-175:0.10)$) {$40^\circ$};
\tikzMarkAngle{(D)}{(A)}{(F)}
\node at ($(D)+(135:0.22)$) {$80^\circ$};
\end{scope}
\end{tikzpicture}
I have searched for many solutions. Finally, no one is satisfactory.
Solution 1: as shown in here, use
\tkzMarkAngle and \tkzLabelAngle
Problem: can't draw an arrow on the arc.
Solution 2: as show in the same place, use
\begin{scope}
\path[clip] (A) -- (E) -- (F) -- cycle;
\draw [red, fill=red!20] (A) circle (10pt);
\end{scope}
Problem: can't draw an arrow on the arc.
Solution 3: the solution in the accepted answer might be complicated for a beginner. And I tried the code, it seems the direction of the arc is not what I expected.
==================================================
My solution: Of course there are many other good solutions. Regarding the problem I am interested in, I came up with a simple solution. I believe many people already know about this. Just post it here for a memo.
%Requires \usetikzlibrary{calc} in the preamble
\begin{tikzpicture}
\coordinate (B) at (2.345,-1.234);
\coordinate (A) at (0,0);
\coordinate (C) at (-2.1,3.2);
\draw [densely dotted, thick] (B)--(A)--(C);
\draw (A) node[below] {$A$};
\draw (B) node[below] {$B$};
\draw (C) node[below left] {$C$};
% label the angle subtended at p1 by p2 and p3
\def\angleRadius{30pt}
% draw the arc
\draw[red,->,thick] let \p1=(A), \p2=(B), \p3=(C),
\n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
($(\p1)!\angleRadius!(\p2)$) arc (\n1:\n2:\angleRadius);
% draw the label
\draw[red] let \p1=(A), \p2=(B), \p3=(C),
\n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
(\p1)+(\n1/2+\n2/2:\angleRadius) node[above] {$\alpha$};
\end{tikzpicture}
Result:
We can use the following code to label the angle which is larger than 180.
\def\angleRadius{30pt}
\draw[red,->,thick] let \p1=(A), \p2=(B), \p3=(C), \n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
($(\p1)!\angleRadius!(\p2)$) arc (\n1:\n2-360:\angleRadius);
\draw[red] let \p1=(A), \p2=(B), \p3=(C), \n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
(\p1)+(\n1/2+\n2/2+180:\angleRadius) node[below] {$\alpha$};
The result is:
Features of the solution:
- Simple. Simply input the three points that determine the angle.
- the arc is generated by arc command. Can set the arrow of the arc and of course other properties.
- Require \usetikzlibrary{calc}, but no \usepackage{tkz-euclide}.
\tkzMarkAngle CAN draw an arrow on the arc! (See: https://tex.stackexchange.com/a/134617/101831)
Jul 19, 2019 at 10:59
You can insert a node at the start of the triangle, at (0,0), and put a label to the left of it with node[left]{$A$}. A full example:
\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[thick]
\draw (0,0) node[left]{$A$} -- (4,0) node[midway,below]{$f$} -- (4,2) node[midway,right]{$\frac{x}{2}$}-- cycle;
\end{tikzpicture}
\end{document}
With tzplot:
\documentclass[tikz]{standalone}
\usepackage{tzplot}
\begin{document}
\begin{tikzpicture}[font=\scriptsize]
% triangle
\tzcoors(0,0)(A){$A$}[180](4,0)(B){$B$}[-45](4,2)(C){$C$}[45];
\tzpolygon(A){$f$}[b](B){$\frac x2$}[r](C){\scshape hypotenuse}[sloped,a](A);
% angle marks
\tzrightanglemark(A)(B)(C){90\textdegree}
\tzanglemark(A)(C)(B){$\theta$}
\tzanglemark(C)(A)(B){$\alpha$}
\tzanglemark'[->,red](C)(A)(B){$\beta$}[pos=1.2](20pt)
\end{tikzpicture}
\end{document}
