6

How to rotate the label counter clockwise such that its base line is parallel to the bisector?

enter image description here

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
    \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
    \pstBissectBAC[linestyle=dotted,linecolor=red]{A}{B}{C}{C'}
    \pstMarkAngle[MarkAngleRadius=1.5,LabelSep=0.75]{A}{B}{C}{\tiny$180^\circ-\theta$}
\end{pspicture}

\end{document}
4
+50
\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
    \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
    \pstBissectBAC[linestyle=dotted,linecolor=red]{A}{B}{C}{B'}
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{}
    \pcline[linestyle=none](B')(B)\ncput[nrot=:U,npos=0.84]{\tiny$180^\circ-\theta$}
\end{pspicture}

\end{document}

a solution without knowing B'

\documentclass{article}
\usepackage{pst-eucl}
\begin{document}    
\begin{pspicture}[showgrid=false](6,6)
    \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{}
  \pcline[linestyle=none](!
    \psGetNodeCenter{A} \psGetNodeCenter{B} \psGetNodeCenter{C} 
    
    /LengthBA B.x A.x sub dup mul B.y A.y sub dup mul add sqrt def
    
    /LengthBC C.x A.x sub dup mul C.y A.y sub dup mul add sqrt def
    
    /Factor LengthBC LengthBA div def
    
    A.x B.x sub Factor mul B.x add /A.x ED
    
    A.y B.y sub Factor mul B.y add /A.y ED
    
    A.x C.x add 2 div A.y C.y add 2 div )(B)\ncput[nrot=:U,npos=0.84]{\tiny$180^\circ-\theta$}

\end{pspicture}

\end{document}
  • It becomes more complicated :-) – kiss my armpit Aug 10 '12 at 14:27
  • no, because you do not need any coodinate – user2478 Aug 10 '12 at 14:30
  • I meant that your solution depends on B' while my solution does not. However, I cannot rotate the label clockwise. – kiss my armpit Aug 10 '12 at 14:35
  • no I do not need the B'. I used it only while it was defined. See edit for another solution – user2478 Aug 10 '12 at 15:15
  • The calculation in the second example will put the label on the median rather than on the angle bisector. – kiss my armpit Aug 10 '12 at 17:49
3

I got the solution even though it is not elegant enough.

enter image description here

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
    \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{\rput{(B)}(0,0){\tiny$180^\circ-\theta$}}
\end{pspicture}

\end{document}

Clockwise rotation:

enter image description here

\documentclass[border=0pt,pstricks]{standalone}
\usepackage{pst-eucl}
\psset{PointName=none,PointSymbol=none}
\begin{document}

\begin{pspicture}[showgrid=false](6,6)
    \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C}
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{\rput{!\psGetNodeCenter{B} B.y B.x atan 180 sub}(0,0){\tiny$180^\circ-\theta$}}
\end{pspicture}

\end{document}

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