# Unit circle combined with angles

I've tried to make a unit circle to explain the angles, but the circle is not on the right place within the defined axes. Am I using the wrong method? Or can't I use the axis definition as in the code. I also want the intersection of the segment and the circle to define sin and cos of the angle (but that I think I will find later with the tkz-euclide)

My code:

\documentclass[11pt,a4paper]{article} % use larger type; default would be 10pt
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all} %% om allerhande objecten te gebruiken zoals gradenboog...
\usetikzlibrary{calc,intersections,through,backgrounds,snakes}
\usepackage{pgfplots}
\pgfplotsset{compat=1.8}
\usepgfplotslibrary{statistics}
\begin{document}
\begin{tikzpicture}
\begin{axis}%
[
grid=major,
x=50mm,
y=50mm,
xmin=-1.1, xmax=1.1,
xtick={-1,0,1},
minor xtick={-1,-0.9,...,1},
xminorgrids = true,
xlabel={\tiny $x$},
axis x line=middle,
ymin=-1.1, ymax=1.1,
ytick={-1,0,1},
minor ytick={-1,-0.9,...,1},
yminorgrids = true,
ylabel={\scriptsize $y$},
axis y line=middle,
no markers,
samples=100,
]

\draw[blue] (axis cs:0,0) circle[radius=1];
\end{axis}
\tkzDefPoint(0,0){A}
\tkzDrawCircle[R](A,5cm)
\tkzDefPoint[shift={(0,0)}](0:5.2){B}
\tkzDefPoint[shift={(0,0)}](50:5.2){C}
\tkzDefPoint[shift={(0,0)}](130:5.2){D}

\tkzDrawSegments[color = red, line width = 1pt](A,B A,C)
\tkzDrawSegments[color = blue, line width = 1pt](A,B A,D)

\tkzDrawPoints(A) \tkzLabelPoints(A)
\tkzMarkAngle[fill= blue,size=2.5cm, opacity=.4](B,A,D);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.7](B,A,C);

\tkzFindAngle(B,A,C) \tkzGetAngle{angleBAC};
\FPround\angleBAC\angleBAC{0}
\tkzLabelAngle[pos = 1](B,A,C){\angleBAC$^\circ$ };

\tkzFindAngle(B,A,D) \tkzGetAngle{angleBAD};
\FPround\angleBAD\angleBAD{0}
\tkzLabelAngle[pos = 2](B,A,D){\angleBAD$^\circ$ };

\end{tikzpicture}
\end{document}


I also have the problem when the angle > 180 it gives the wrong angle, because \tkzGetAngle only works in the interval -180° +180°. ## 1 Answer

You sort of ask three quite different questions. For the first:

By default the anchor of a pgfplots axis is set to south west, and the position is set to (0,0) in the coordinate system of tikzpicture. You can change the position with at={(x,y)}, but as your circle is set around (0,0) that isn't necessary. You just need to add anchor=center to the axis options.

For the second:

To get the intersection between the line segments and the circle you can use

\tkzInterLC[R](A,C)(A,5cm)\tkzGetSecondPoint{CC}
\tkzInterLC[R](A,D)(A,5cm)\tkzGetSecondPoint{DC}


CC and DC is the intersections. In the below code I've drawn and labeled those points, but I haven't drawn the lines corresponding to the sine and cosine. \documentclass[11pt]{standalone} % use larger type; default would be 10pt
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all} %% om allerhande objecten te gebruiken zoals gradenboog...
\usetikzlibrary{calc,intersections,through,backgrounds,snakes}
\usepackage{pgfplots}
\pgfplotsset{compat=1.8}
\usepgfplotslibrary{statistics}
\begin{document}
\begin{tikzpicture}
\begin{axis}%
[
anchor=center,  % sets axis anchor to the axis origin
grid=major,
x=50mm,
y=50mm,
xmin=-1.1, xmax=1.1,
xtick={-1,0,1},
minor xtick={-1,-0.9,...,1},
xminorgrids = true,
xlabel={\tiny $x$},
axis x line=middle,
ymin=-1.1, ymax=1.1,
ytick={-1,0,1},
minor ytick={-1,-0.9,...,1},
yminorgrids = true,
ylabel={\scriptsize $y$},
axis y line=middle,
no markers,
samples=100,
]

\draw[blue] (axis cs:0,0) circle[radius=1];
\end{axis}
\tkzDefPoint(0,0){A}
\tkzDrawCircle[R](A,5cm)
\tkzDefPoint[shift={(0,0)}](0:5.2){B}
\tkzDefPoint[shift={(0,0)}](50:5.2){C}
\tkzDefPoint[shift={(0,0)}](130:5.2){D}

\tkzDrawSegments[color = red, line width = 1pt](A,B A,C)
\tkzDrawSegments[color = blue, line width = 1pt](A,B A,D)

% Finds the intersections of segments and circle
\tkzInterLC[R](A,C)(A,5cm)\tkzGetSecondPoint{CC}
\tkzInterLC[R](A,D)(A,5cm)\tkzGetSecondPoint{DC}

% draw and label points
\tkzDrawPoints(A,CC,DC) \tkzLabelPoints(A,CC,DC)

\tkzMarkAngle[fill= blue,size=2.5cm, opacity=.4](B,A,D);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.7](B,A,C);

\tkzFindAngle(B,A,C) \tkzGetAngle{angleBAC};
\FPround\angleBAC\angleBAC{0}
\tkzLabelAngle[pos = 1](B,A,C){\angleBAC$^\circ$ };

\tkzFindAngle(B,A,D) \tkzGetAngle{angleBAD};
\FPround\angleBAD\angleBAD{0}
\tkzLabelAngle[pos = 2](B,A,D){\angleBAD$^\circ$ };

\end{tikzpicture}
\end{document}

• +1. Very pretty solution. I's typing one using \begin{scope}[xshift=5.5cm,yshift=5.5cm]\end{scope} to affect the second part of the code. – Sigur Dec 30 '13 at 10:51
• @Sigur don't understand your comment. Where should I put your code? As Torbjørn T. code is working for me. – Arne Timperman Dec 30 '13 at 12:42
• @Arne, don't worry. I was saying that my solution was not so pretty. You don't need to use my code. – Sigur Dec 30 '13 at 13:00