# Drawing circular arcs of 60 degrees in TikZ

I am trying to draw circular arcs at two angles of measure 60 degrees. Why don't

\draw[draw=green] ($(B')!0.375cm!(A')$) arc (180:120:0.375);
\draw[draw=blue] ($(P')!3.75mm!(B')$) arc (0:60:0.375);


render circular arcs? A', P', and B' are all points on a horizontal line.

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage[dvipsnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}

\begin{tikzpicture}[x=0.3cm, y=0.3cm]

\path let \n1={6.5*(sqrt(3)/2)} in (0,0) coordinate (A) (10,0) coordinate (B) (6.75,\n1) coordinate (C) (3.25,\n1) coordinate (D);
\path[fill=LimeGreen!75] (A) -- (C) -- (D) -- cycle;
\path[fill=LimeGreen!75] (B) -- (C) -- (6.75,0) -- cycle;
\path[fill=NavyBlue!50] (A) -- (C) -- (6.75,0) -- cycle;
\draw[OliveGreen] (A) -- (D) -- (C);
\draw[OliveGreen] (B) -- (C) -- (6.75,0);
\draw[OliveGreen] (B) -- (6.75,0);
\draw[blue] (A) -- (C);
\draw[blue] (A) -- (6.75,0);
\draw[dashed, thick, yellow] (6.75,0) -- (C);

%An arrow from the given isosceles trapezoid to a decomposition of it into a parallelogram and two congruent right triangles.
\draw[-latex] let \n1={6.5*(sqrt(3)/2)/2} in (11,\n1) to[out=60,in=120] (13,\n1);

\path let \n1={6.5*(sqrt(3)/2)} in (14,0) coordinate (A') (24,0) coordinate (B') (20.75,\n1) coordinate (C') (17.25,\n1) coordinate (D')
(17.5,0) coordinate (P');
\draw[OliveGreen] (A') -- (D') -- (C');
\draw[OliveGreen] (B') -- (C') -- (17.5,0);
\draw[OliveGreen] (B') -- (20.75,0);
\path[fill=LimeGreen!75] (A') -- (C') -- (D') -- cycle;
\path[fill=NavyBlue!50] (A') -- (C') -- (P') -- cycle;
\draw[blue] (A') -- (20.75,0);
\draw[blue, dashed] (C') -- (P');
\draw[blue] (A') -- (C');

%The legs of the isosceles trapezoid and BP are marked with "|".
\draw[OliveGreen, thick] ($($(A')!0.5!(D')$)!3pt!90:(A')$) -- ($($(A')!0.5!(D')$)!3pt!-90:(A')$);
\draw[OliveGreen, thick] ($($(B')!0.5!(C')$)!3pt!90:(B')$) -- ($($(B')!0.5!(C')$)!3pt!-90:(B')$);
\draw[blue, thick] ($($(C')!0.5!(P')$)!3pt!90:(C')$) -- ($($(C')!0.5!(P')$)!3pt!-90:(C')$);
\draw[dashed, thick, yellow] (20.75,0) -- (C');

%The marks indicating the measure of \angle{ABC} and \angle{BPC} are drawn. Since they are congruent to each other, they are marked with "|".
\draw[draw=green] ($(B')!0.375cm!(A')$) arc (180:120:0.375);
\draw[draw=green] ($(B') +(150:{0.375cm-3pt})$) -- ($(B') +(150:{0.375cm+3pt})$);
%
\draw[draw=blue] ($(P')!3.75mm!(B')$) arc (0:60:0.375);
\draw[draw=blue] ($(P') +(30:{0.375cm-3pt})$) -- ($(P') +(30:{0.375cm+3pt})$);

\end{tikzpicture}

\end{document}

• This is bizarre. Why are the arc commands the only commands in this code that are not properly implemented? – A gal named Desire May 11 at 22:23
• I just hope that someone else will have a look at this. Since I am very confident that my TeX installation works, for me the most likely explanations are (i) you compile a different code than the above or (ii) something is wrong with your TeX installation. – user121799 May 11 at 22:33
• This is so stupid - I edited my code replacing (10,0) coordinate (B) with ({(3/10)*10},0) coordinate (B) and making similar modifications with the other coordinates ... and I get exactly what I want! – A gal named Desire May 12 at 14:15
• Please use @<username> when replying to specific users, otherwise they won't be notified (only the user who wrote the question/answer is notified of all comments). Also, I suggest you edit the code in your question and add [x=0.3cm, y=0.3cm], so that the question actually makes sense. I do get misplaced arcs when adding those options. – Torbjørn T. May 12 at 14:21
• I got i.stack.imgur.com/lB2oA.png, which doesn't look like what you're describing. Edit: try \begin{tikzpicture}[scale=0.3] – Torbjørn T. May 12 at 14:23

Specify the unit of the arc radius:

\draw[draw=green] ($(B')!0.375cm!(A')$) arc (180:120:0.375cm);

\draw[draw=blue] ($(P')!3.75mm!(B')$) arc (0:60:0.375cm); (By the way, what you were seeing are circular arcs, but the radius is smaller.)

• Isn't the default in centimeters? – A gal named Desire May 12 at 14:34
• Did the options in the tikzpicture change the 0.375 in \draw[draw=green] ($(B')!0.375cm!(A')$) arc (180:120:0.375cm); to (3/10)*0.375? – A gal named Desire May 12 at 14:35
• Haven't tried figuring out exactly what happens, but could be that if no unit is specified, that it is scaled by the unit vector, so when you change the unit vectors to be 0.3cm instead of 1cm, the radius becomes 0.375*0.3cm. – Torbjørn T. May 12 at 14:36