标记三角形中的角度

Question

我想让 TikZ 在笛卡尔平面上绘制一个三角形 - 一个顶点位于原点 O、A = (2,1) 和 B = (-3, 5) 的三角形。我还想绘制并标记两个角度 - 一个从正 x 轴到 OA，一个从正 x 轴到 OB。我希望角度在与 OA 和 OB 相切的地方有箭头。我还想在序言中保留“\documentclass{amsart}”。

Answer 1

使用两种简单的可能性TikZ：

1. 对于版本 3.0，使用angles和quotes库：

   

   代码：

   \documentclass{article}
   \usepackage{tikz}
   \usetikzlibrary{angles,quotes}
   
   \tikzset{
   mydot/.style={
     fill,
     circle,
     inner sep=1.5pt
     }
   }
   
   \begin{document}
   
   \begin{tikzpicture}[>=latex]
   % the coordinates of the vertices
   \coordinate (O) at (0,0);
   \coordinate (A) at (2,1);
   \coordinate (B) at (-3,5);
   \coordinate (E) at (2,0);
   
   % the axis
   \draw[help lines,->] (-3.5,0) -- (2.5,0);
   \draw[help lines,->] (0,-0.5) -- (0,5.5);
   
   % the edges of the triangle    
   \draw (O) -- (A) -- (B) -- cycle;
   
   % labelling the vertices
   \node[mydot,label={right:$A$}] at (A) {};
   \node[mydot,label={left:$B$}] at (B) {};
   \node[mydot,label={below:$O$}] at (O) {};
   
   % the arcs for the angles
   \path[gray]
     pic["$\alpha$" shift={(23pt,3pt)},draw,->,angle radius=1.5cm] {angle = E--O--A}
     pic["$\beta$" above=6pt,draw,->,angle radius=0.75cm] {angle = E--O--B};
   \end{tikzpicture}
   
   \end{document}
   

2. 对于版本 2.10，不带库并使用arc路径：

   

   代码：

   \documentclass{amsart}
   \usepackage{tikz}
   
   \tikzset{
   mydot/.style={
     fill,
     circle,
     inner sep=1.5pt
     }
   }
   
   \begin{document}
   
   \begin{tikzpicture}[>=latex]
   % the coordinates of the vertices
   \coordinate (O) at (0,0);
   \coordinate (A) at (2,1);
   \coordinate (B) at (-3,5);
   
   % the axis
   \draw[help lines,->] (-3.5,0) -- (2.5,0);
   \draw[help lines,->] (0,-0.5) -- (0,5.5);
   
   % the edges of the triangle    
   \draw (O) -- (A) -- (B) -- cycle;
   
   % labelling the vertices
   \node[mydot,label={right:$A$}] at (A) {};
   \node[mydot,label={left:$B$}] at (B) {};
   \node[mydot,label={below:$O$}] at (O) {};
   
   % the arcs for the angles    
   \begin{scope}[gray]
   \draw[->] 
     (1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=41] node[midway,right] {$\alpha$};
   \draw[->] 
     (0.5,0) +(0:0.25cm) arc [radius=0.75cm,start angle=0,end angle=122] node[midway,above] {$\beta$};
   \end{scope}
   \end{tikzpicture}
   
   \end{document}