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);

• 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