[Tex/LaTex] Shift Arc Location –> Reference Angle

Question

I'm making worksheets to help students remember the unit circle and/or special triangles.

I want the blue arrow to represent the reference angle, between the terminal arm and the x-axis.

How do I do that?

\documentclass{article}

\usepackage{tikz}
 \newcommand{\MyScale}{1.75}

\newcommand{\OneFifty}
{%
Sketch \(\theta=150^\circ\) in standard position, labeling all sides, the angle, and the reference angle. Then, state all its trigonometric ratios.

\begin{tikzpicture}[scale=\MyScale]
\draw[thick,<->] (-2,0)--(1,0); % x-axis
\draw[thick,<->] (0,-1)--(0,1.5); % y-axis
\draw (0,0)--(-1.732,1); % terminal arm
\draw (-1.732,1)--(-1.732,0); % vertical drop
\draw (-1.532,0)--(-1.532,0.2)--(-1.732,0.2); %right-angle mark
\node[anchor=center] at (-0.866,-0.2) {A \(=-\sqrt{3}\)};
\node[anchor=center] at (-2.1,0.5) {OP \(=1\)};
\node[anchor=center] at (-0.8, 0.8) {H \(=2\)};
%Angle in standard position.
\node[red, anchor=center] at (1.4,0.3) {\(\theta=150^\circ\)};
\draw[red,->] (0.866,0) arc (0:150:0.866); % Angle in SP
\node[blue, anchor=center] at (-0.3,0.2) {\(\theta_{r} =30^{\circ}\)}; % Reference Angle label
\draw[blue,<->] (0.35,0) arc (150:180:0.35); % Reference Angle
\end{tikzpicture}%

\(\sin{150^\circ}=\frac{1}{2}\)

\(\cos{150^\circ}=\frac{-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}\)

\(\tan{150^\circ}=\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}\)

\(\csc{150^\circ}=\frac{2}{1}=2\)

\(\sec{150^\circ}=\frac{2}{-\sqrt{3}}=-\frac{2\sqrt{3}}{3}\)

\(\cot{150^\circ}=\frac{-\sqrt{3}}{1}=-\sqrt{3}\)
}

\begin{document}

\OneFifty

\end{document}

Answer 1

This can be implemented elegantly using the angles library as suggested by @TorbjørnT. You need also quotes library to label angles. I changed all \( .. \) to $ .. $; there is nothing wrong with parentheses but it hurts my eyes and I prefer the more distinct symbol $. Finally, I don't think a \newcommand is very beneficial here as this figure is not likely repeated many times. You may think of removing the newcommand and inserting the figure as well as the text directly.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}   % <<----
\newcommand{\MyScale}{1.75}
\newcommand{\OneFifty}
{%
Sketch $\theta=150^\circ$ in standard position, labeling all sides, the angle, and the reference angle. Then, state all its trigonometric ratios.

\begin{tikzpicture}[scale=\MyScale]
\draw[thick,<->] (-2,0)coordinate(B)--(1,0)coordinate(C); % x-axis <<----
\draw[thick,<->] (0,-1)--(0,1.5); % y-axis
\draw (0,0)coordinate(O)--(-1.732,1)coordinate(A); % terminal arm  <<----
\draw (-1.732,1)--(-1.732,0); % vertical drop
\draw (-1.532,0)--(-1.532,0.2)--(-1.732,0.2); %right-angle mark
\node[anchor=center] at (-0.866,-0.2) {A \(=-\sqrt{3}\)};
\node[anchor=center] at (-2.1,0.5) {OP \(=1\)};
\node[anchor=center] at (-0.8, 0.8) {H \(=2\)};
% Angle in standard position.

\pic [draw,<->, blue, angle radius=8mm, "$\theta_{r} =30^\circ$", angle eccentricity=2] {angle=A--O--B};
\pic [draw,<->, red, angle radius=8mm, "$\theta=150^\circ$", angle eccentricity=1.4] {angle=C--O--A};

\end{tikzpicture}%

$\sin{150^\circ}=\frac{1}{2}$

$\cos{150^\circ}=\frac{-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}$

$\tan{150^\circ}=\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}$

$\csc{150^\circ}=\frac{2}{1}=2$

$\sec{150^\circ}=\frac{2}{-\sqrt{3}}=-\frac{2\sqrt{3}}{3}$

$\cot{150^\circ}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$
}

\begin{document}

\OneFifty

\end{document}