# Drawing an arc tangent to a line

Thanks to a guy in this forum I was able to handle several triangles I had to draw. Now a smaller problem. I did the triangle in the picture, but I just miss the arch tangent to the point. Every time I adjust the center, it moves and I had to start from the beginning. I just need the arc and the letter omega and D2.

Here's what I've done so far:

\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}


Like this?

\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]{$D_2$};
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}


For the future, I'd kindly ask you to post complete minimal working examples since otherwise only those who know by looking at the code which libraries are needed can answer the question immediately.

• yes thank you, but I need the arc to be smaller, like from 90 to 60 degrees like in the picture Apr 23 '20 at 22:15
• @Noob_Latex The angle is 30 degrees, as can be seen in arc[start angle=90,end angle=120,radius=4.5]. You can change the value of the end angle to whatever suits you, e.g, 105.
– user194703
Apr 23 '20 at 22:17
• if I move the angle if I move to angle like this: \draw[thick] (0,-9) arc[start angle=120,end angle=60,radius=5] the arc moves as if I moved the center. Apr 23 '20 at 22:24
• @Noob_Latex I think I understand the question better now, and updated the answer accordingly.
– user194703
Apr 23 '20 at 22:27
• thanks it works. And yes I will post a minimal code, sorry for that. Apr 23 '20 at 22:47