A mistake using the arc syntax or the let-in syntax

Question

In the display, I have three angles that have the same measure. (These are the only three angles that are marked. They are centered at points O, P, and Q.) I am using this code to get familiar with the arc syntax and the let-in syntax. To draw the arc at P, I declare the starting point to be ($(P)!0.5cm!(R)$). This part of the command is correct. I am not sure the reason that the arc is not drawn about P, though.

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

\usepackage{tikz}
\usetikzlibrary{calc,positioning,intersections,quotes}



\begin{document}

\vfill
\pagebreak

\noindent \hspace*{\fill}
\begin{tikzpicture}

\coordinate (O) at (0,0);

\draw[-latex,name path=ray_1] (O) -- (10:7);
\coordinate (label_for_ray_r_1) at ($(10:7) +(10:3mm)$);
\node at (label_for_ray_r_1){$r_{1}$};
\node at ($(O)!-3mm!(10:7)$){$O$};
\draw[-latex,name path=ray_2] (O) -- (135:3.5);
\coordinate (label_for_ray_r_2) at ($(135:3.5) +(135:0.3)$);
\node at (label_for_ray_r_2){$r_{2}$};
\draw[-latex,name path=ray_3] (O) -- (342.5:8);
\coordinate (label_for_ray_r_3) at ($(342.5:8) +(342.5:0.3)$);
\node at (label_for_ray_r_3){$r_{3}$};

%The ray opposite to $r_{2}$ is drawn.
\draw[-latex,dashed,name path=opposite_to_ray_2] (O) -- (-45:8);


\coordinate (Q) at (342.5:6.5);
\draw[fill] (Q) circle (1.5pt);
\coordinate (A) at ($(O)!(Q)!(10:7)$);
\node at ($(A)!-3mm!(Q)$){$A$};
\draw[fill,blue] (A) circle (1.5pt);
\draw[name path=path_AQ] (A) -- (Q);


%The projection of Q onto the ray opposite $r_{2}$ is called $P$.
\coordinate (P) at ($(O)!(Q)!(-45:7)$);
\node at ($(P)!3mm!90:(O)$){$P$};
\node at ($(Q)!-3mm!(P)$){$Q$};
\draw[dashed] (Q) -- (P);

%The projection of P onto $r_{1}$ is called $B$.
\coordinate (B) at ($(O)!(P)!(10:7)$);
\node at ($(B)!3mm!-90:(O)$){$B$};
\draw[dashed] (B) -- (P);


%The intersection of the line through P with the same slope as $r_{1}$ and the line through A and Q
%is labeled $R$.
\coordinate (a_point_R_1_in_the_plane_to_determine_R) at ($(P) +(10:3)$);
\path[name path=P_to_R_1] (P) -- (a_point_R_1_in_the_plane_to_determine_R);
\coordinate (a_point_R_2_in_the_plane_to_determine_R) at ($(Q)!-1!(A)$);
\path[name path=P_to_R_2] (Q) -- (a_point_R_2_in_the_plane_to_determine_R);
\coordinate[name intersections={of=P_to_R_1 and P_to_R_2, by={R}}];
\draw[fill] (R) circle (1.5pt);
\draw[dashed] (P) -- (R);
\draw[dashed] (Q) -- (R);

%The label for R is typeset.
\coordinate (label_R_below) at ($(R)!-7mm!(Q)$);
\coordinate (label_R_right) at ($(R)!-7mm!(P)$);
\coordinate (label_R) at ($(label_R_below)!0.5!(label_R_right)$);
\node[blue] at ($(R)!3mm!(label_R)$){$R$};


%An angle is drawn at O.
\draw[draw=blue] (O) ++(10:5mm) arc (10:-45:5mm);


%An angle at P  with measure x is drawn.
\draw[draw=blue] let \p1=($(B)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(Q)-(P)$), \n2={atan(\y2/\x2)} in ($(P)!0.5cm!(B)$) arc (\n1:\n2:0.5);


%An angle at Q  with measure x is drawn.
\draw[draw=blue] let \p1=($(Q)-(P)$), \n1={atan(\y1/\x1)}, \p2=($(Q)-(R)$), \n2={atan(\y2/\x2)} in ($(Q)!0.5cm!(P)$) arc (\n1:\n2:0.5);


\end{tikzpicture}

\end{document}

Answer 1

The problem comes from the fact that atan give answer modulo 180° because it doesn't take acount of quadrant, use atan2 instead. You may think use macro instead on repetiting code. Once it works, make a macro.

On can compare atan in red, atan2 in blue : the second computes itself the right non oriented angle but both are not sentive to trignometric orientation.

In green : solves the two problems.

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

\usepackage{tikz}
\usetikzlibrary{calc,positioning,intersections,quotes}


\newcommand{\AngleBad}[4][red]{%
\draw[#1] let
    \p1=($(#2)-(#3)$),
    \n1={atan(\y1/\x1)},
    \p2=($(#4)-(#3)$),
    \n2={atan(\y2/\x2)} in
    ($(#3)!.4cm!(#2)$) arc (\n1:\n2:.4) ;
}

\newcommand{\AngleBetter}[4][blue]{%
\draw[#1] let
    \p1=($(#2)-(#3)$),
    \n1={atan2(\y1,\x1)},
    \p2=($(#4)-(#3)$),
    \n2={atan2(\y2,\x2)} in
    ($(#3)!.5cm!(#2)$) arc (\n1:\n2:.5) ;
}

\newcommand{\MarkAngle}[4][green]{%
    \pgfmathanglebetweenpoints{\pgfpointanchor{#3}{center}}{\pgfpointanchor{#2}{center}}%
    \let\AngleA\pgfmathresult
    \pgfmathanglebetweenpoints{\pgfpointanchor{#3}{center}}{\pgfpointanchor{#4}{center}}%
    \let\AngleB\pgfmathresult
    \pgfmathparse{ifthenelse(\AngleA>\AngleB,\AngleA-360,\AngleA)}
    \let\AngleA\pgfmathresult ; 
    \draw[#1] ($(#3)!.6cm!(#2)$) arc (\AngleA:\AngleB:.6) ; }


\begin{document}

\vfill
\pagebreak

\noindent \hspace*{\fill}
\begin{tikzpicture}

\coordinate (O) at (0,0);

\draw[-latex,name path=ray_1] (O) -- (10:7);
\coordinate (label_for_ray_r_1) at ($(10:7) +(10:3mm)$);
\node at (label_for_ray_r_1){$r_{1}$};
\node at ($(O)!-3mm!(10:7)$){$O$};
\draw[-latex,name path=ray_2] (O) -- (135:3.5);
\coordinate (label_for_ray_r_2) at ($(135:3.5) +(135:0.3)$);
\node at (label_for_ray_r_2){$r_{2}$};
\draw[-latex,name path=ray_3] (O) -- (342.5:8);
\coordinate (label_for_ray_r_3) at ($(342.5:8) +(342.5:0.3)$);
\node at (label_for_ray_r_3){$r_{3}$};

%The ray opposite to $r_{2}$ is drawn.
\draw[-latex,dashed,name path=opposite_to_ray_2] (O) -- (-45:8);


\coordinate (Q) at (342.5:6.5);
\draw[fill] (Q) circle (1.5pt);
\coordinate (A) at ($(O)!(Q)!(10:7)$);
\node at ($(A)!-3mm!(Q)$){$A$};
\draw[fill,blue] (A) circle (1.5pt);
\draw[name path=path_AQ] (A) -- (Q);


%The projection of Q onto the ray opposite $r_{2}$ is called $P$.
\coordinate (P) at ($(O)!(Q)!(-45:7)$);
\node at ($(P)!3mm!90:(O)$){$P$};
\node at ($(Q)!-3mm!(P)$){$Q$};
\draw[dashed] (Q) -- (P);

%The projection of P onto $r_{1}$ is called $B$.
\coordinate (B) at ($(O)!(P)!(10:7)$);
\node at ($(B)!3mm!-90:(O)$){$B$};
\draw[dashed] (B) -- (P);


%The intersection of the line through P with the same slope as $r_{1}$ and the line through A and Q
%is labeled $R$.
\coordinate (a_point_R_1_in_the_plane_to_determine_R) at ($(P) +(10:3)$);
\path[name path=P_to_R_1] (P) -- (a_point_R_1_in_the_plane_to_determine_R);
\coordinate (a_point_R_2_in_the_plane_to_determine_R) at ($(Q)!-1!(A)$);
\path[name path=P_to_R_2] (Q) -- (a_point_R_2_in_the_plane_to_determine_R);
\coordinate[name intersections={of=P_to_R_1 and P_to_R_2, by={R}}];
\draw[fill] (R) circle (1.5pt);
\draw[dashed] (P) -- (R);
\draw[dashed] (Q) -- (R);

%The label for R is typeset.
\coordinate (label_R_below) at ($(R)!-7mm!(Q)$);
\coordinate (label_R_right) at ($(R)!-7mm!(P)$);
\coordinate (label_R) at ($(label_R_below)!0.5!(label_R_right)$);
\node[blue] at ($(R)!3mm!(label_R)$){$R$};


\AngleBad{P}{O}{B}
\AngleBad{B}{O}{P} % not sensitive to trigonometric orientation
\AngleBad{P}{Q}{R}
\AngleBad{Q}{P}{B}

\AngleBetter{P}{O}{B}
\AngleBetter{B}{O}{P} % not sensitive to trigonometric orientation
\AngleBetter{P}{Q}{R}
\AngleBetter{Q}{P}{B}

\MarkAngle{P}{O}{B} % sensitive to trigonometric orientation
\MarkAngle[dashed,green]{B}{O}{P}
\MarkAngle{P}{Q}{R}
\MarkAngle{Q}{P}{B}


\end{tikzpicture}

\end{document}

Answer 2

I admire elegance of Tarass answer, but OP challenge me to show solution without use of atan. In this solution you need to know the quadrant where is angle and accordingly add or subtracts 180 degrees. O.K, here we go:

\documentclass[border=1mm,
               class=amsart,
               preview]{standalone}
\usepackage{amsmath,amsfonts}

\usepackage{tikz}
\usetikzlibrary{calc,positioning,intersections,quotes}

    \begin{document}
\begin{tikzpicture}
\coordinate (O) at (0,0);

\draw[-latex,name path=ray_1] (O) -- (10:7);
\coordinate (label_for_ray_r_1) at ($(10:7) +(10:3mm)$);
\node at (label_for_ray_r_1){$r_{1}$};
\node at ($(O)!-3mm!(10:7)$){$O$};
\draw[-latex,name path=ray_2] (O) -- (135:3.5);
\coordinate (label_for_ray_r_2) at ($(135:3.5) +(135:0.3)$);
\node at (label_for_ray_r_2){$r_{2}$};
\draw[-latex,name path=ray_3] (O) -- (342.5:8);
\coordinate (label_for_ray_r_3) at ($(342.5:8) +(342.5:0.3)$);
\node at (label_for_ray_r_3){$r_{3}$};

%The ray opposite to $r_{2}$ is drawn.
\draw[-latex,dashed,name path=opposite_to_ray_2] (O) -- (-45:8);

\coordinate (Q) at (342.5:6.5);
\draw[fill] (Q) circle (1.5pt);
\coordinate (A) at ($(O)!(Q)!(10:7)$);
\node at ($(A)!-3mm!(Q)$){$A$};
\draw[fill,blue] (A) circle (1.5pt);
\draw[name path=path_AQ] (A) -- (Q);

%The projection of Q onto the ray opposite $r_{2}$ is called $P$.
\coordinate (P) at ($(O)!(Q)!(-45:7)$);
\node at ($(P)!3mm!90:(O)$){$P$};
\node at ($(Q)!-3mm!(P)$){$Q$};
\draw[dashed] (Q) -- (P);

%The projection of P onto $r_{1}$ is called $B$.
\coordinate (B) at ($(O)!(P)!(10:7)$);
\node at ($(B)!3mm!-90:(O)$){$B$};
\draw[dashed] (B) -- (P);

%The intersection of the line through P with the same slope as $r_{1}$ and the line through A and Q is labeled $R$.
\coordinate (a_point_R_1_in_the_plane_to_determine_R) at ($(P) +(10:3)$);
\path[name path=P_to_R_1] (P) -- (a_point_R_1_in_the_plane_to_determine_R);
\coordinate (a_point_R_2_in_the_plane_to_determine_R) at ($(Q)!-1!(A)$);
\path[name path=P_to_R_2] (Q) -- (a_point_R_2_in_the_plane_to_determine_R);
\coordinate[name intersections={of=P_to_R_1 and P_to_R_2, by={R}}];
\draw[fill] (R) circle (1.5pt);
\draw[dashed] (P) -- (R);
\draw[dashed] (Q) -- (R);

%The label for R is typeset.
\coordinate (label_R_below) at ($(R)!-7mm!(Q)$);
\coordinate (label_R_right) at ($(R)!-7mm!(P)$);
\coordinate (label_R) at ($(label_R_below)!0.5!(label_R_right)$);
\node[blue] at ($(R)!3mm!(label_R)$){$R$};

%Angles at O, P and Q:
\path[draw=blue] 
%An angle is drawn at O.
    (O) ++(10:5mm) arc (10:-45:5mm)
%An angle at P  with measure x is drawn.
    let \p1=($(B)-(P)$), 
        \n1={atan(\y1/\x1)}, 
        \p2=($(Q)-(P)$), 
        \n2={atan(\y2/\x2)} in 
    ($(P)!0.5cm!(B)$) arc (\n1+180:\n2:0.5)% added 180 degree to \n1
%An angle at Q  with measure x is drawn.
    let \p1=($(Q)-(P)$), 
        \n1={atan(\y1/\x1)}, 
        \p2=($(Q)-(R)$),
        \n2={atan(\y2/\x2)} in
    ($(Q)!0.5cm!(P)$) arc (\n1-180:\n2:0.5);% substracted 180 degree from \n1
\end{tikzpicture}
    \end{document}

And obtained picture:

Answer 3

Here's a version in Metapost for comparison. In particular I think the drawing is simplified by the use of rotated and angle which were deliberately designed to remove the need for you to think explicitly about sines, cosines, and arc-tangents. (See the Metafont Book, p.67).

prologues := 3;
outputtemplate := "%j%c.eps";

beginfig(1);

% define the end points of the three rays
z1 = right scaled 200 rotated 10;
z2 = right scaled 100 rotated 135;
z3 = right scaled 225 rotated -17.5;

% define the other points, relative to Q
pair A, B, P, Q, R;
Q = 0.8125 z3;
A = whatever[origin, z1]; A-Q = whatever * z1 rotated 90;
P = whatever[origin, z2]; P-Q = whatever * z2 rotated 90;
B = whatever[origin, z1]; B-P = whatever * z1 rotated 90;
R = whatever[A,Q];        R-P = whatever * (B-P) rotated 90;

% mark the angles 
drawoptions(withcolor .67 blue);
draw fullcircle scaled 30 rotated angle (Q-P) shifted P cutafter (P--B);
draw fullcircle scaled 30 rotated angle (P-Q) shifted Q cutafter (Q--R);
draw fullcircle scaled 30 rotated angle P               cutafter (origin--z1);
drawoptions();

% draw the rays and A--Q
drawarrow origin -- z1; label(btex $r_1$ etex, z1 scaled 1.05);
drawarrow origin -- z2; label(btex $r_2$ etex, z2 scaled 1.08);
drawarrow origin -- z3; label(btex $r_3$ etex, z3 scaled 1.05);
draw A--Q;

% draw the dashed lines
drawoptions(dashed evenly);
draw B--P--R--Q--P;
drawarrow origin -- P scaled 4/3;
drawoptions();

% label the points
dotlabel.urt(btex $Q$ etex, Q);
dotlabel.top(btex $A$ etex, A);
dotlabel.lrt(btex $R$ etex, R) withcolor .67 blue;

label.top (btex $B$ etex, B);
label.llft(btex $P$ etex, P);
label.llft(btex $O$ etex, origin);

endfig;
end.

Note: the construction P = whatever[a,b] defines P as a point that lies somewhere on the line through a and b. While P-Q = whatever * (a-b) says that "the direction of P from Q should be some multiple of the direction of a from b". You can, of course, omit b when it is the origin.

Plain MP defines origin=(0,0) for you.