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繪製與直線相切的圓弧

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繪製與直線相切的圓弧

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感謝這個論壇中的一個人，我能夠處理我必須繪製的幾個三角形。現在是一個較小的問題。我做了圖中的三角形，但我只是錯過了該點的拱切線。每次我調整中心時，它都會移動，我必須從頭開始。我只需要弧線、字母 omega 和 D2。

這是我到目前為止所做的：

\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

像這樣？

\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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對於未來，我懇請您發布完整的最小工作範例，否則只有那些透過查看程式碼知道需要哪些函式庫的人才能立即回答問題。

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