# Drawing arcs with radius not at origin

I'm looking for the best way (meaning simplest) to draw an arc for the interior angle of a triangle (or any other regular polygon). I'm using a triangle presently. Here's my minimal working example:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{figure}
\centering
\begin{tikzpicture}
\node[draw, minimum size=3cm,regular polygon,regular polygon sides=3] (a) {};
\draw (a.corner 1) arc  (240:300:0.5cm)node[below]{$\alpha = 120\degree$};
%\draw (a.corner 1) -- node[right]{$h$} (a.south) node[below]{$b$};
\draw (0,0) circle (1.5cm);
\end{tikzpicture}
\caption{Regular Triangle; arcs inscribed}
\label{fgr:inscribedtri}
\end{figure}
\end{document}


This is what it produces

This is a bit closer to what I'm looking for.

As a neophyte of tikz (and little more in LaTeX) I'm really in need of assistance). Of course, I don't what the alpha there and alpha isn't equal to 120 degrees (that is a remnant of some past stuff that isn't what I wanted). I want that arc that is terribly drawn in the top of the triangle with GIMP. I want a node below it reading "alpha = 60 degrees".

As I mentioned, my attempt is above. Is it simpler to place a vertex of the triangle at the origin and work it from there? I don't really care. Thanks.

• If you have TikZ > v3.0.0 search for angles library in the manual – percusse Oct 30 '16 at 9:07

You need to define right starting point for arc.

Edit: Starting point of arc lie on line (a.corner 1) -- (a.corner 2), which cross a.corner 1 with angle of 240 degree. In this direction on distance of 1.2 cm is starting point of arc. This is calculated in ($(a.corner 1)+(240:1.2)$):

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles, calc, quotes, shapes.geometric}
\usepackage{siunitx}

\begin{document}
\begin{figure}
\centering
\begin{tikzpicture}
\node[draw, minimum size=3cm,regular polygon,regular polygon sides=3] (a) {};
\draw (0,0) circle (1.5cm);
%
\draw ($(a.corner 1)+(240:1.2)$) % arc start point
arc (240:300:1.2)
node[midway,below] {$\alpha = \SI{120}{\degree}$};
\end{tikzpicture}
\hfil
\begin{tikzpicture}[% with library angles
my angle/.style = {draw,
angle eccentricity=1.3,
font=\large} % angle label position!
]
\node (a) [draw, minimum size=3cm,
regular polygon, regular polygon sides=3] {};
\draw (0,0) circle (1.5cm);
%
\coordinate (A) at (a.corner 1);
\coordinate (B) at (a.corner 2);
\coordinate (C) at (a.corner 3);
%
\path   pic [my angle, "$\alpha = \SI{120}{\degree}$"] {angle = B--A--C};

\end{tikzpicture}
\caption{Regular Triangle; arcs inscribed}
\label{fgr:inscribedtri}
\end{figure}
\end{document}


Edit: image on the right side is drawn by help of tikz library angles and quotes.

• Thank you much. Can you explain to me why, for the left triangle, the beginning coordinates for the arc are enclosed in '$' \draw ($(a.corner 1)+(240:1.2)$)? – Andrew Falanga Oct 31 '16 at 1:24 • As I said, you need to define starting point of arc it lie on line (a.corner 1) -- (a.corner 2), which cross a.corner 1 at angle 240 degree. In this direction on distance 1.2 cm is starting point of starting point. This is calculated in ($(a.corner 1)+(240:1.2)$). – Zarko Oct 31 '16 at 1:30 • You need to pint @AndrewFalanga if the conversation involves more than one person other than the answerer. So I got pinged but AF won't have. – cfr Oct 31 '16 at 2:56 • Good to know. For future. – Zarko Oct 31 '16 at 9:51 • @AndrewFalanga, for the case if you not notice my last comment: starting point of arc lie on line (a.corner 1) -- (a.corner 2), which cross a.corner 1 with angle of 240 degree. In this direction on distance 1.2 cm is starting point of arc. This is calculated in ($(a.corner 1)+(240:1.2)\$)`. I will add this to the answer. – Zarko Oct 31 '16 at 9:56