[Tex/LaTex] Marking angles in a triangle

tikz-pgf

I would like to have TikZ draw a triangle on the Cartesian plane – a triangle with vertices at the origin O, A = (2,1), and B = (-3, 5). I would also like to have two angles drawn and labeled – one from the positive x-axis to OA and one from the positive x-axis to OB. I would like the angles to have arrows where they touch OA and OB. I would also like to keep "\documentclass{amsart}" in the preamble.

Best Answer

Two simple possibilities using TikZ:

For version 3.0, using the angles and quotes libraries:

enter image description here

The code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (A) at (2,1);
\coordinate (B) at (-3,5);
\coordinate (E) at (2,0);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle    
\draw (O) -- (A) -- (B) -- cycle;

% labelling the vertices
\node[mydot,label={right:$A$}] at (A) {};
\node[mydot,label={left:$B$}] at (B) {};
\node[mydot,label={below:$O$}] at (O) {};

% the arcs for the angles
\path[gray]
  pic["$\alpha$" shift={(23pt,3pt)},draw,->,angle radius=1.5cm] {angle = E--O--A}
  pic["$\beta$" above=6pt,draw,->,angle radius=0.75cm] {angle = E--O--B};
\end{tikzpicture}

\end{document}

For version 2.10, without libraries and using the arc path:

enter image description here

The code:

\documentclass{amsart}
\usepackage{tikz}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (A) at (2,1);
\coordinate (B) at (-3,5);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle    
\draw (O) -- (A) -- (B) -- cycle;

% labelling the vertices
\node[mydot,label={right:$A$}] at (A) {};
\node[mydot,label={left:$B$}] at (B) {};
\node[mydot,label={below:$O$}] at (O) {};

% the arcs for the angles    
\begin{scope}[gray]
\draw[->] 
  (1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=41] node[midway,right] {$\alpha$};
\draw[->] 
  (0.5,0) +(0:0.25cm) arc [radius=0.75cm,start angle=0,end angle=122] node[midway,above] {$\beta$};
\end{scope}
\end{tikzpicture}

\end{document}

Related Solutions

[Tex/LaTex] Picture indicating the fifth roots of unity

To learn more about TikZ read pgfmanual.pdf. Via the macro \n you can adjust the order of the root to be taken.

\documentclass[tikz]{standalone}
\usepackage{amsmath}% for \Re and \Im
\def\n{5}
\begin{document}
\begin{tikzpicture}[
  dot/.style={draw,fill,circle,inner sep=1pt}
  ]
  \draw[->] (-2,0) -- (2,0) node[below] {$\Re$};
  \draw[->] (0,-2) -- (0,2) node[left] {$\Im$};
  \draw[help lines] (0,0) circle (1);

  \node[dot,label={below right:$O$}] (O) at (0,0) {};
  \foreach \i in {1,...,\n} {
    \node[dot,label={\i*360/\n-(\i==\n)*45:$w_{\i}$}] (w\i) at (\i*360/\n:1) {};
    \draw[->] (O) -- (w\i);
  }
  \draw[->] (0:.3) arc (0:360/\n:.3);
  \node at (360/\n/2:.5) {$\alpha$};
\end{tikzpicture}
\end{document}

For 1 < n < 13:

[Tex/LaTex] Labeling sides and angles of a right triangle for an argument of the Pythagorean Theorem

Here's one possibility using "pure" TikZ:

enter image description here

The image was produced using simply

\begin{tikzpicture}
\RectTri{(0,3)}{(1,0)}{6cm}
\begin{scope}[xshift=8.5cm]
\RectTri[black]{(0,0)}{(4,2)}{4cm}
\end{scope}
\end{tikzpicture}

\RectTri has three mandatory arguments; the first two are the coordinates for the vertices of one of the legs and the third one is the length of the second leg. The optional argument lets you customize the style used to draw the triangle.

The code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,angles,quotes,decorations.markings}

\newcommand\RectTri[4][thick,green!50!black,text=black]{%
\coordinate [label=left:$C$] (C) at #2;
\coordinate [label=below right:$B$] (B) at #3;
\coordinate (aux) at ($ #2 ! 1 ! 90:#3 $);
\coordinate [label=above:$A$] (A) at ($ #2 !#4!(aux) $);

\coordinate (perp) at ($(A)!(C)!(B)$);
\draw[purple!70!black,thick,dashed] (C) -- (perp);

\draw[#1] 
  (C) -- 
  node[auto] {$b$} (A) -- 
  node[auto] {$c$} (B) --
  node[auto] {$a$} 
  (C)
  pic ["$\alpha$",draw,cyan,thick,angle radius=1cm] {angle = C--A--B} 
  pic ["$\alpha$",draw,cyan,thick,angle radius=1cm] {angle = B--C--perp}
  pic ["$\beta$",draw,orange!70!black,thick,angle radius=1cm] {angle = A--B--C}
  pic ["$\beta$",draw,orange!70!black,thick,angle radius=1cm] {angle = perp--C--A};
}

\begin{document}

\begin{tikzpicture}
\RectTri{(0,3)}{(1,0)}{6cm}
\begin{scope}[xshift=8.5cm]
\RectTri[black]{(0,0)}{(4,2)}{4cm}
\end{scope}
\end{tikzpicture}

\end{document}

And here's an approach using tkz-euclide:

\documentclass{article}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}
\tkzDefPoint(0,1){A}
\tkzDefPoint(2,4){C}
\tkzDefPointWith[orthogonal normed,K=7](C,A)
\tkzGetPoint{B}

\tkzLabelPoint[left](A){$A$}
\tkzLabelPoint[right](B){$B$}
\tkzLabelPoint[above](C){$C$}

\tkzMarkRightAngle(A,C,B)

\tkzDrawSegment[green!60!black](A,C)
\tkzDrawSegment[green!60!black](C,B)
\tkzDrawSegment[green!60!black](B,A)

\tkzLabelSegment[auto](B,A){$c$}
\tkzLabelSegment[auto,swap](B,C){$a$}
\tkzLabelSegment[auto,swap](C,A){$b$}

\tkzDrawAltitude[dashed,color=magenta](A,B)(C)
\tkzGetPoint{D}

\tkzMarkAngle[size=1cm,color=cyan,mark=|](C,B,A)
\tkzMarkAngle[size=1cm,color=cyan,mark=|](A,C,D)
\tkzMarkAngle[size=0.75cm,color=orange,mark=||](D,C,B)
\tkzMarkAngle[size=0.75cm,color=orange,mark=||](B,A,C)
\end{tikzpicture}

\end{document}

enter image description here

Best Answer

Related Solutions

[Tex/LaTex] Picture indicating the fifth roots of unity

[Tex/LaTex] Labeling sides and angles of a right triangle for an argument of the Pythagorean Theorem

Related Question