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绘制以不同于原点的点为中心的圆弧

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绘制以不同于原点的点为中心的圆弧

我在原点画一个 30 度角。组成这个角的两条线段的公共端点是，D其他端点是E'和F。 F距离4线段DE'。我想画一个以 F 为中心，半径为 的圆弧5。这个圆弧应该DE'在两点相交。我不知道用我的代码画的是什么圆弧！

制作直角标记的代码有什么错误？

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{tkz-euclide}
\usetkzobj{all}


% Refer to http://tex.stackexchange.com/questions/196181/why-do-my-arcs-end-at-the-wrong-node-positions
% Refer to  http://www.texample.net/tikz/examples/intersecting-lines/
% To have TikZ calculate the foot of the altitude of a triangle, see Chapter 13, section 5 in the TikZ manual.


\begin{document}



\noindent {\bf 19.) }In $\triangle{\mathit{DEF}}$,  $\vert DF \vert = 8$, $\vert EF \vert = 5$, and $\angle{\mathit{EDF}} = 30$. How many such triangles are there?
\vskip0.2in

\noindent \hspace*{\fill}
\begin{tikzpicture}

%These commands draw edges $DF$ and $E'F$ so that $\angle{\mathit{E'DF}} = 30$.
%$DF$ is to be drawn at an angle of 120 degrees and DE' is to be drawn at an
%angle of angle of 150 degrees. Since the length of DE' is not specified, it
%is to be drawn as a ray.
\coordinate (D) at (0,0);
\node (vertex_D) at ($(D) + (-45:7.5pt)$){$D$};
\coordinate (F) at (120:8);
\node (vertex_F) at ($(F) + (90:7.5pt)$){$F$};
\coordinate (E') at (150:12);
\node (vertex_E') at ($(E') + (-90:7.5pt)$){$E'$};
\draw (D) -- (F);
\draw (D) -- (E');

\path pic[draw, angle radius=10mm,"$30$",angle eccentricity=1.25] {angle = F--D--E'};

%These commands label the length of DF.
\coordinate (DF_midpoint) at ($(D)!0.5!(F)$);
\node (DF_midpoint_label) at ($(DF_midpoint) + (30:7.5pt)$) {8};


%These commands draw the height of the triangle from F to line segment $DE'$.
%The height is labeled $h$.
\coordinate (P) at ($(D)!(F)!(E')$);
\draw [dashed] (F) -- (P) node [midway, right]{$h = 4$};
\node (point_P) at ($(P) + (-120:7.5pt)$){$P$};


%These commands draw an arc of a circle centered at F with radius 5.
\draw[dashed,draw=blue!30] (F) arc (150:290:5);

%The following commands make the right-angle mark and "colors" the inside of it white.
\coordinate (U) at ($(P)!4mm!45:(D)$);
\draw (U) -- ($(D)!(U)!(E')$);
\draw (U) -- ($(F)!(U)!(P)$);

\end{tikzpicture}
\hspace{\fill}
\vskip0.25in

\end{document}

答案1

圆弧没有中心指定，因此您有一个起始/终止角度和半径。而且几何形状不正确。

首先，F 位于 D 的 120 度线上，因此您必须放置一个相对于该点的点。在这里，我还移动到 120 度线上的一个点，并从 120 度开始绘制圆弧。如果我移动到其他地方，则需要考虑。

我还注释掉了导致直角出现问题的那行。在

毕达哥拉斯定理论证图

我已经用不同的语法修复了这个问题。

\begin{tikzpicture}

\coordinate[label=-45:$D$] (D) at (0,0);
\coordinate[label=90:$F$] (F) at (120:8);
\coordinate[label=-90:$E$] (E) at (150:12);
\draw (F) -- (D) -- (E);
\path pic[draw, angle radius=10mm,"$30$",angle eccentricity=1.25] {angle = F--D--E};
\coordinate[label=45:$8$] (DF_midpoint) at ($(D)!0.5!(F)$);
\coordinate[label=-135:$P$] (P) at ($(D)!(F)!(E)$);
\draw [dashed] (F) -- (P) node [midway, right]{$h = 4$};

\draw[dashed,draw=blue!30] (F) ++(120:5) arc (120:290:5);

\coordinate (U) at ($(P)!4mm!45:(D)$);
%\draw (U) -- ($(D)!(U)!(E)$); % Not necessarily parallel to FP !!!
\draw (U) -- ($(F)!(U)!(P)$);
\end{tikzpicture}

在此处输入图片描述

答案2

由于您正在加载tkz-euclide包，因此您可以指定中心（ tikzarc命令不指定中心），使用

\tkzDrawArc[R, color=blue,dashed,thick](F, 5cm)(135,315)
  • F是中心。
  • 5cm是半径。
  • 最后两个数字分别是起始和终止的极坐标。
  • 最后，R是一个选项（来自 fr.人造丝，半径），您可以指定中心，然后指定半径。

输出

图1

代码

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{tkz-euclide}
\usetkzobj{all}


% Refer to http://tex.stackexchange.com/questions/196181/why-do-my-arcs-end-at-the-wrong-node-positions
% Refer to  http://www.texample.net/tikz/examples/intersecting-lines/
% To have TikZ calculate the foot of the altitude of a triangle, see Chapter 13, section 5 in the TikZ manual.


\begin{document}
\begin{tikzpicture}

%These commands draw edges $DF$ and $E'F$ so that $\angle{\mathit{E'DF}} = 30$.
%$DF$ is to be drawn at an angle of 120 degrees and DE' is to be drawn at an
%angle of angle of 150 degrees. Since the length of DE' is not specified, it
%is to be drawn as a ray.
\coordinate (D) at (0,0);
\node (vertex_D) at ($(D) + (-45:7.5pt)$){$D$};
\coordinate (F) at (120:8);
\node (vertex_F) at ($(F) + (90:7.5pt)$){$F$};
\coordinate (E') at (150:12);
\node (vertex_E') at ($(E') + (-90:7.5pt)$){$E'$};
\draw (D) -- (F);
\draw (D) -- (E');

\path pic[draw, angle radius=10mm,"$30$",angle eccentricity=1.25] {angle = F--D--E'};

%These commands label the length of DF.
\coordinate (DF_midpoint) at ($(D)!0.5!(F)$);
\node (DF_midpoint_label) at ($(DF_midpoint) + (30:7.5pt)$) {8};


%These commands draw the height of the triangle from F to line segment $DE'$.
%The height is labeled $h$.
\coordinate (P) at ($(D)!(F)!(E')$);
\draw [dashed] (F) -- (P) node [midway, right]{$h = 4$};
\node (point_P) at ($(P) + (-120:7.5pt)$){$P$};

%These commands draw an arc of a circle centered at F with radius 5.
%\draw[dashed,draw=blue!30] (F) arc (150:290:5);
\tkzDrawArc[R,color=blue,dashed, thick](F, 5)(125,315)

%The following commands make the right-angle mark and "colors" the inside of it white.
\coordinate (U) at ($(P)!4mm!45:(D)$);
\draw (U) -- ($(D)!(U)!(E')$);
\draw (U) -- ($(F)!(U)!(P)$);

\end{tikzpicture}
\hspace{\fill}
\vskip0.25in

\end{document}

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