Ich verzweifel noch an dieser angle library...

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Code, hier editierbar zum Übersetzen:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,quotes,intersections}
\tikzset{
angle radius=0.75cm
}
\begin{document}
\section*{Scheitelwinkel}
\begin{tikzpicture}
\draw[name path=A] (0,0) coordinate (A) -- (5,0) coordinate (B);
\draw[name path=B] (1,2) coordinate (C) -- (3,-1) coordinate (D);
\path[name intersections={of=A and B,name=i}];
\draw pic["$\alpha$", draw=red, ->] {angle=D--i-1--B};
\draw pic["$\alpha'$", draw=blue, ->] {angle=C--i-1--A};
\end{tikzpicture}
\section*{Stufenwinkel}
\begin{tikzpicture}
\draw[name path=A] (0,0) coordinate (A) -- (5,0) coordinate (B);
\draw[name path=B] (3,2) coordinate (C) -- (1,-1) coordinate (D);
\draw[name path=C] (0,1) coordinate (E) -- (5,1) coordinate (F);
\path[name intersections={of=A and C,name=i}];
\path[name intersections={of=B and C,name=j}];
\draw pic["$\alpha$", draw=red, ->] {angle=B--i-1--C};
\draw pic["$\alpha'$", draw=blue, ->] {angle=C--j-1--E};
\end{tikzpicture}
\end{document}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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Wieso schließt der Winkel im Stufenwinkelschaubild nicht?

gefragt 05 Jun '16, 04:19

runix's gravatar image

runix
1.0k257586
Akzeptiert-Rate: 0%

bearbeitet 05 Jun '16, 09:31

Bitte immer nur eine Frage auf einmal.

(05 Jun '16, 08:41) saputello

Du musst die richtigen Linien für den Schnittpunkt auswählen. Im zweiten Bild haben A und C nämlich keinen Schnittpunkt.

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\path[name intersections={of=A and C,name=i}];
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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Daher wird die Koordinate (i-1) nicht überschrieben sondern aus dem vorhergehenden Bild verwendet (deshalb kompiliert auch dein Beispiel in dieser Frage nicht). Ändert man es so, dass die korrekten Linien verwendet werden, nämlich A und B

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\path[name intersections={of=A and B,name=i}];
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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tut es wunderbar. Ich habe die Zeile unten noch mit einem Pfeil markiert.

Open in Online-Editor
Code, hier editierbar zum Übersetzen:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,quotes,intersections}
\tikzset{
angle radius=0.75cm
}
\begin{document}
\section*{Scheitelwinkel}
\begin{tikzpicture}
\draw[name path=A] (0,0) coordinate (A) -- (5,0) coordinate (B);
\draw[name path=B] (1,2) coordinate (C) -- (3,-1) coordinate (D);
\path[name intersections={of=A and B,name=i}];
\draw pic["$\alpha$", draw=red, ->] {angle=D--i-1--B};
\draw pic["$\alpha'$", draw=blue, ->] {angle=C--i-1--A};
\end{tikzpicture}
\section*{Stufenwinkel}
\begin{tikzpicture}
\draw[name path=A] (0,0) coordinate (A) -- (5,0) coordinate (B);
\draw[name path=B] (3,2) coordinate (C) -- (1,-1) coordinate (D);
\draw[name path=C] (0,1) coordinate (E) -- (5,1) coordinate (F);
\path[name intersections={of=A and B,name=i}]; % <--
\path[name intersections={of=B and C,name=j}];
\draw pic["$\alpha$", draw=red, ->] {angle=B--i-1--C};
\draw pic["$\alpha'$", draw=blue, ->] {angle=C--j-1--E};
\end{tikzpicture}
\end{document}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
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alt text

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beantwortet 05 Jun '16, 10:50

Henri's gravatar image

Henri
15.7k133943
Akzeptiert-Rate: 46%

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