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绘制与直线相切的圆弧

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绘制与直线相切的圆弧

在此处输入图片描述

感谢论坛上的一位朋友，我能够处理好必须绘制的几个三角形。现在问题变小了。我画了图中的三角形，但只是错过了与点相切的拱。每次我调整中心时，它都会移动，我不得不从头开始。我只需要弧线和字母 omega 和 D2。

以下是我目前所做的：

\begin{figure}[h]
  \centering
  \begin{tikzpicture}[>=latex]

    \coordinate (A) at (0,0);
    \coordinate (B) at (7,0);
    \coordinate (C) at (5,-4);
    \coordinate (ABmid) at (A -| C);
    \coordinate (D) at (0,-4);
    \coordinate (E) at (10,-4);
    \coordinate (F) at (2,0);

    \draw[->] [thick] (A)--(B) node[above,midway]{$U_2$};
    \draw[->] [thick] (C)--(A) node[left,midway]{$W_2$};
    \draw[->] [thick] (C)--(B) node[right,midway]{$C_2$};
    \draw [thick,dashed] (D)--(E);
    \draw[->] [thick] ($(B)+(0,5pt)$)--($(ABmid) +(0,5pt)$) node[above,pos=0.5]{$C_{\theta2}$};
    \draw[thick,dashed] (C)--(F);
    \draw[->] [thick] (C)--(ABmid) node[left,pos=0.7]{$C_{a2}$};

    \pic["$\beta_{2b}$",draw, angle eccentricity=1.2,angle radius=15mm] {angle = F--C--D};
    \pic["$\beta_2$",draw, angle eccentricity=1.2,angle radius=25mm] {angle = A--C--D}; 
    \pic["$\alpha_2$",draw, angle eccentricity=1.2,angle radius=12mm] {angle = E--C--B};

    \draw[thick] (0,-9) arc (180:0:5);

  \end{tikzpicture}
  \caption{\textit{Triangoli di velocità all'estremità della girante}}
  \label{fig:Triangoli girante}
\end{figure}

答案1

像这样？

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,angles,quotes} 
\begin{document}
\begin{figure}[h]
  \centering
  \begin{tikzpicture}[>=latex]

    \coordinate (A) at (0,0);
    \coordinate (B) at (7,0);
    \coordinate (C) at (5,-4);
    \coordinate (ABmid) at (A -| C);
    \coordinate (D) at (0,-4);
    \coordinate (E) at (10,-4);
    \coordinate (F) at (2,0);

    \draw[->] [thick] (A)--(B) node[above,midway]{$U_2$};
    \draw[->] [thick] (C)--(A) node[left,midway]{$W_2$};
    \draw[->] [thick] (C)--(B) node[right,midway]{$C_2$};
    \draw [thick,dashed] (D)--(E);
    \draw[->] [thick] ($(B)+(0,5pt)$)--($(ABmid) +(0,5pt)$) node[above,pos=0.5]{$C_{\theta2}$};
    \draw[thick,dashed] (C)--(F);
    \draw[->] [thick] (C)--(ABmid) node[left,pos=0.7]{$C_{a2}$};

    \pic["$\beta_{2b}$",draw, angle eccentricity=1.2,angle radius=15mm] {angle = F--C--D};
    \pic["$\beta_2$",draw, angle eccentricity=1.2,angle radius=25mm] {angle = A--C--D}; 
    \pic["$\alpha_2$",draw, angle eccentricity=1.2,angle radius=12mm] {angle = E--C--B};

    \draw[thick] ($(C)+(60:5)-(90:5)$) arc[start angle=60,end angle=120,radius=5]
    coordinate[pos=0.25] (p);
    \draw[<-] (p) -- ++ (-180+75:1) node[anchor=75]{$D_2$};
    \draw[<-] ([yshift=-0.5cm]C) arc[start angle=90,end angle=110,radius=4.5]
    node[midway,below=1ex]{$\omega$};

  \end{tikzpicture}
  \caption{\textit{Triangoli di velocit\`a all'estremit\`a della girante.}}
  \label{fig:Triangoli girante}
\end{figure}
\end{document}

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对于未来，我恳请您发布完整的最小工作示例，否则只有那些通过查看代码知道需要哪些库的人才能立即回答这个问题。

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