我如何才能准确定位节点锚点？

Question

所以我有以下 tikz 代码：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
    \coordinate (O) at (0,0);
    \path (O) -- ++(4,0) coordinate (B);
    \path[name path=l1] (O) -- ++(40:10);
    \path[name path=l2] (B) -- ++(120:10);
    
    \draw [name intersections={of=l1 and l2, by=A}]
    (O) -- (A) node[midway, above left] {$a$} -- (B) -- cycle node[midway, below] {$b$};
    
    \draw (O) ++(0.5,0) arc [start angle=0, end angle=40, radius=0.5] node[above right, midway] {$\theta$};
    \node[left] at (O) {$O$};
    \node[above] at (A) {$A$};
    \node[right] at (B) {$B$};
\end{tikzpicture}
\end{document}

这里，theta 标签被 OA 线覆盖。这是个问题，因为看起来不太好。我想把它放低一点，这样它就不再被覆盖了。但是，我可以在节点上放置的唯一其他“方向”是right，这将导致 theta 标签被 OB 线覆盖。我该如何设置方向或锚点，使其现在恰好位于两条线之间？

Answer 1

我做了两件事：我改为above right，right并且我在节点\strut中添加了\theta。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
    \coordinate (O) at (0,0);
    \path (O) -- ++(4,0) coordinate (B);
    \path[name path=l1] (O) -- ++(40:10);
    \path[name path=l2] (B) -- ++(120:10);
    
    \draw [name intersections={of=l1 and l2, by=A}]
    (O) -- (A) node[midway, above left] {$a$} -- (B) -- cycle node[midway, below] {$b$};
    
    \draw (O) ++(0.5,0) arc [start angle=0, end angle=40, radius=0.5] 
      node[right,midway] {$\theta$\strut};
    \node[left] at (O) {$O$};
    \node[above] at (A) {$A$};
    \node[right] at (B) {$B$};
\end{tikzpicture}
\end{document}

如果你想要更多的控制，只需添加一个底部延伸\rule到\theta“推它”，就像

\draw (O) ++(0.5,0) arc [start angle=0, end angle=40, radius=0.5] 
   node[right,midway] {$\theta$\rule[-6pt]{0pt}{0pt}};