# Drawing a triangle with smooth curves in the middle

I need to draw a picture like this.

I made it by using Paint so it is not very clear. I need the smooth curves inside the triangle as well as the straight lines that connect the curves and the circle to be dashed. The symbol that appears two times is \epsilon.

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There's a circular sector shape that might help you. – Claudio Fiandrino Jan 30 '14 at 13:13

Run with xelatex or latex->dvips->ps2pdf

\documentclass{article}
\begin{document}

\psset{unit = 2}
\begin{pspicture}(-1.7,-1.7)(2,2)
\psaxes[xlabelsep =-0.02,xlabelOffset = 0.07,
ylabelsep = -0.02,ylabelOffset = 0.1,
arrowscale=1.5,linewidth=0.5pt]{->}(0,0)(-1.7,-1.7)(2,2)
\pscircle(0,0){1}
\psset{origin={0.15,0.05}}
\pswedge[fillcolor=black!10,fillstyle=solid](0,0){0.65}{0}{30}
\pscustom[fillcolor=black!10,fillstyle=solid]{
\psline(1.05;30)(1.5;30)(1.5,0)
\psarc(0,0){1.05}{0}{30}}
\pcline(0.8;20)(1.2;20)
\naput[npos=0.2]{\small $\epsilon$}
\naput[npos=0.8]{\small $\epsilon$}
\naput[npos=1.5,labelsep=0.2]{\small $\overline{T}$}
\end{pspicture}

\end{document}


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Thanks too much ... thats wonderful!! – Ruzayqat Jan 30 '14 at 15:13
The best PSTricks solution, as usual. :) – Svend Tveskæg Jan 30 '14 at 15:22

This is a solution where the starting angle=15 deg (with radius=1.5cm) and ending angle=45 deg for the pie inside. As to the sector region, the radii are 2.5cm and 4cm respectively. These all can be changed to suit one's need.

The syntax for drawing an arc can be done via

\draw (x,y) arc (alph:beta:r);     % where x+jy=r\angle alph
\draw (alph:r) arc (alph:beta:r);  % start angle=alpha, end angle=beta, radius=r.


Code

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning,intersections}

\begin{document}

\begin{tikzpicture}
\draw (-2.5,0)node[left]{-1}--(2.5,0) node[right]{1};
\draw (0,2.5) node[right]{1}--(0,-2.5)node[right]{-1};
\draw (0,0) circle (2cm);
\draw (0,0) -- +(15:1.5cm) arc (15:45:1.5cm) -- cycle;
\draw (15:2.5)--(15:4) arc (15:45:4) --node[left]{$\bar T$} (45:2.5) arc (45:15:2.5);
\draw (25:1.5)--node[pos=0.2,above]{$\epsilon$}
node[pos=0.7,above]{$\epsilon$}(25:2.5);
\end{tikzpicture}

\end{document}

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Thanks for the help.. many thanks – Ruzayqat Jan 30 '14 at 15:14

Just another solution with PSTricks.

\documentclass[pstricks,border=12pt,12pt]{standalone}
\usepackage{pst-eucl}
\psset{unit=2,PointSymbol=none,linejoin=1}
\begin{document}
\multido{\i=15+5}{15}{%
\begin{pspicture}(-1.5,-1.5)(2,1.5)
\pstGeonode[PosAngle={-135,-45,-135,135},PointName={-1,1,-1,1}](-1,0){A}(1,0){B}(0,-1){C}(0,1){D}
\pcline[nodesep=-.5](A)(B)
\pcline[nodesep=-.5](C)(D)
\pscircle{1}
\def\temp{%
\pnodes(0,0){O}(1.5;10){P}(1.5;\i){Q}%
\pswedge{.5}{(P)}{(Q)}%
\pscustom{\psarc(O){1}{(P)}{(Q)}\psline(Q)(P)\closepath}%
\pstBissectBAC[linestyle=none,PointName=none]{P}{O}{Q}{R}%
\pnodes([nodesep=.5]{R}O){X}([nodesep=1]{R}O){Y}%
\pcline(X)(Y)%
\psset{labelsep=2pt}
\naput[npos=.2]{$\epsilon$}%
\naput[npos=.8]{$\epsilon$}%
\pcline[linestyle=none]([nodesep=1]{Q}O)(Q)%
\naput{$\overline{T}$}%
}%
\rput(.3;10){\temp}
\end{pspicture}}

\end{document}


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A PSTricks solution:

\documentclass{article}

\usepackage{expl3}
\ExplSyntaxOn
\cs_new_eq:NN \calc \fp_eval:n
\ExplSyntaxOff

\def\xOrigin{0.15 }
\def\yOrigin{0.05 }

\begin{document}

\psset{unit = 2}
\begin{pspicture}(-1.7,-1.7)(2,2)
\psaxes[
xlabelsep = -0.02,
xlabelOffset = 0.07,
ylabelsep = -0.02,
ylabelOffset = 0.1
]{->}(0,0)(-1.7,-1.7)(2,2)
\pscircle(0,0){1}
\psarc(\xOrigin,\yOrigin){1.05}{0}{30}
{\psset{origin = {\xOrigin,\yOrigin}}
{\psset{linestyle = none, offset = 4pt}
\pcline(0.8;20)(1;20)
\ncput{\small $\epsilon$}
\pcline(1;20)(1.2;20)
\ncput{\small $\epsilon$}
\ncput{\small $\overline{T}$}}}