# Drawing angle addition in latex

I'm trying to draw the picture below but have been unsuccessful (I drew part a, can't do part b)

This is what I used to draw part a

\begin{center}
\begin{tikzpicture}[font=\boldmath]\large
\coordinate (A) at (0,0) {};
\coordinate (B) at (2,0) {};
\coordinate (C) at (1,1) {};
\coordinate (D) at (3,1) {};

\begin{scope}[shift={(0,1.5)}]
\draw[thick, black, dotted]  (4,0) -- (5,1) node[sloped,midway,above] {};
\draw[thick, black, dotted]   (1,1) -- (5,1) node[sloped,midway,above] {};
\draw[->,  thick, black,   arrows={-latex}]  (0,0) -- (4,0) node[sloped,right=-0.7cm,above] {$\mathsf{A}$};
\draw[->,  thick, black,  arrows={-latex}]  (0,0) -- (1,1) node[sloped,midway,above=-0.1cm] {$\mathsf{B}$};
\draw[->,  thick, black, arrows={-latex}]  (0,0) -- (5,1) node[sloped,midway,above=-0.1cm] {$\mathsf{A+B}$};
\end{scope}
\end{tikzpicture}
\end{center}


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Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. – yo' Jul 31 '14 at 22:23

One possibility:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{angles,decorations.pathreplacing}

\begin{document}

\begin{tikzpicture}[
>=latex]
\draw[dashed]
(5,0) coordinate (X) --
(0,0) coordinate (O);

\path
coordinate (A) at (20:4.5cm)
coordinate (B) at (55:3cm)
coordinate (AB) at (75:5cm);

\foreach \Coor in {A,B,AB}
{
\draw[->,thick,shorten >= 2pt] (O) -- (\Coor);
\node[circle,fill,inner sep=1.5pt,label={above right:$\Coor$}]
at (\Coor) {};
}
\node[draw,circle,fill=white,inner sep=1.5pt,label={below left:$O$}] {};

\draw[decorate,decoration={brace,raise=5pt},black!70,text=black]
(O) --
node[sloped,above=5pt] {(length $A$)$\cdot$(length $B$)}
(AB);

\path[->,black!70]
\end{tikzpicture}

\end{document}


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I wonder if determining of coordinates A, B and AB as given in your answer do have some advantages over just saying: \coordinate (A) at (...);\coordinate (B) at (...);\coordinate (AB) at (...);? – Zarko Aug 1 '14 at 14:35

Another possibility could be:

\documentclass[tikz,border=0.1cm]{standalone}
\usetikzlibrary{decorations.pathreplacing}
\begin{document}
\begin{tikzpicture}[font=\boldmath, >=latex, decoration = {brace,raise=4pt},line width=0.7pt]
\foreach \n/\s/\l/\p/\a in {o/draw/{$0$}/{0,0}/left,A/fill/{$A$}/{20:5}/right,B/fill/{$B$}/{45:3.5}/right,AB/fill/{$AB$}/{65:6}/right}
\node (\n)[\s,circle,scale=0.3,label={\a:\l}]at (\p){};
\draw (o)  edge[dashed] ++(5,0)
[->] edge (A)
[->] edge (B)
[->] edge (AB);
\draw[decorate,thin] (o) -- node[sloped,above=5pt] {(length $A$)$\cdot$(length $B$)} (AB);
\foreach \t/\a/\b/\r in {/0/20/1.6,thin/20/65/1.6,thin/0/45/2,/45/65/2}
\draw [->,\t](\a:\r) arc (\a:\b:\r);
\end{tikzpicture}
\end{document}


With the output:

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