# Tikz arc angle calculation

I am having a serious mental block this afternoon. I have the following code to generate a shape, and I need one more arc to close it and cycle back to the original point. I can't seem to get my mind around it for some reason. Any help would be much appreciated.

\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{positioning}

\begin{document}
\begin{tikzpicture}

\coordinate (dm1) at (0,0);
\coordinate$right=1.5cm of dm1$ (dm2);

\draw$thin$ (dm1) -- (dm2)
arc (0:60:1cm)
arc (0:180:0.3cm)
;

\end{tikzpicture}
\end{document}

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Why do you have $right=..$, and \draw$thin$? –  Torbjørn T. Aug 14 '14 at 20:31
Please stop posting code including spurious characters. I'm not sure what is causing this but it makes it more difficult for people to help since they cannot cut, paste and compile your code. –  cfr Aug 14 '14 at 21:03

Obviously it is hard to know how the shape should be completed since you do not say but here is one possibility:

\documentclass[tikz]{standalone}
\usetikzlibrary{positioning}

\begin{document}
\begin{tikzpicture}

\coordinate (dm1) at (0,0);
\coordinate [right=1.5cm of dm1] (dm2);

\draw [thin] (dm1) -- (dm2)
arc (0:60:1cm)
arc (0:180:0.3cm)
arc (120:180:1cm)
-- cycle
;

\end{tikzpicture}
\end{document}

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Thank you, that is exactly what I wanted. I had a strange mental block this afternoon... Much appreciated. –  ozsu Aug 14 '14 at 22:23
@ozsu Note that cycle completes the base of the figure since 1.5cm is not sufficient. –  cfr Aug 14 '14 at 23:39