这是我目前所拥有的:
\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{tikz}
\begin{document}
\begin{center}
\begin{tikzpicture}
\coordinate (S) at (0,0); %does nothing
\draw (0,0) circle (2cm);
\draw (-2,0) -- (2,0);
\end{tikzpicture}
\end{center}
\end{document}
抱歉,我是 tikz 的新手,所以很抱歉我的问题比较基础。有人能帮忙吗?
答案1
这是一个解决方案pstricks
:
\documentclass[border=10pt, svgnames]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(-2,-2)(2,2)
\psset{linejoin=1}
\pnodes(-2,0){A}(2,0){B}(2;120){C}(0,0){S} \uput[dr](S){$S$}
\psset{LabelSep=0.4, MarkAngleRadius=0.6, linecolor=LightSeaGreen}
\everypsbox{\small\color{LightSeaGreen}}
\pstMarkAngle{S}{A}{C}{$\color{LightSeaGreen}\alpha$}
\pstMarkAngle{A}{C}{S}{$\color{LightSeaGreen}\alpha$}
\psset{linecolor=DeepSkyBlue, MarkAngleRadius=0.9, LabelSep=0.7}
\everypsbox{\small\color{DeepSkyBlue}}
\pstMarkAngle{S}{C}{B}{$\beta$}
\pstMarkAngle{C}{B}{S}{$\beta$}
\psset{linecolor=black}
\pstCircleAB{A}{B}
\psline(A)(C)(B)(A)\psline(C)(S)
\end{pspicture}
\end{document}
答案2
您可以只使用TikZ
(参见 Zarko 的回答) 来绘制它,或者如果您有大量这样的工作 (欧几里得几何),则使用 tkz-euclide。而且您可以混合tkz-euclide
使用TikZ
。
\documentclass[margin=.5cm]{standalone}
\usepackage{tkz-euclide}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}
\tikzmath{ \r=6 ;}
\tkzDefPoint(0,0){O}
\tkzDefPoint(0:\r){B}
\tkzDefPoint(150:\r){C}
\tkzDefPoint(180:\r){A}
\tkzDrawPolygon(A,B,C)
\tkzDrawSegment(O,C)
\tkzDrawPoints(A,B,C,O)
\tkzDrawCircle(O,A)
\tkzLabelAngles[pos=.5](B,A,C A,C,O){$\alpha$}
\tkzLabelAngles[pos=1.6](C,B,O O,C,B){$\beta$}
\tkzMarkAngles[mark=|](B,A,C A,C,O)
\tkzMarkAngles[mark=||,size=2](C,B,O O,C,B)
\tkzLabelPoints(A,B)
\tkzLabelPoints[above](C)
\end{tikzpicture}
\end{document}
使用下一个代码,您将得到一个 3-4-5 直角三角形(埃及或毕达哥拉斯)
\documentclass[margin=.5cm]{standalone}
\usepackage{tkz-euclide}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}[rotate=36.9,scale=2]
\tkzDefPoint(0,0){A}
\tkzDefPoint(4,0){C}
\tkzDefTriangle[egyptian](A,C) \tkzGetPoint{B}
\tkzDrawPolygon(A,B,C)
\tkzDefMidPoint(A,B) \tkzGetPoint{O}
\tkzDrawSemiCircle(O,B)
\tkzDrawSegment(O,C)
\tkzLabelPoints(O,A,B)
\tkzLabelPoints[above](C)
\tkzMarkRightAngle[fill=teal!20,opacity=.4](B,C,A)
\tkzLabelAngles[pos=.75](B,A,C A,C,O){$\alpha$}
\tkzLabelAngles[pos=.75](C,B,O O,C,B){$\beta$}
\tkzMarkAngles[mark=|](B,A,C A,C,O)
\tkzMarkAngles[mark=||](C,B,O O,C,B)
\end{tikzpicture}
\end{document}
如果你想要 A 的一个特定角度(这里是 30 度)
\documentclass[margin=.5cm]{standalone}
\usepackage{tkz-euclide}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}
\tkzInit[xmin=-6,xmax=6,ymin=-1,ymax=6]
\tkzGrid
\tikzmath{\r = 6 ;}
\tkzDefPoint(-\r,0){A}
\tkzDefPoint(\r,0){B}
\tkzDefPoint(0,0){O}
\tkzDefShiftPoint[A](30:2){c}
\tkzInterLC(A,c)(O,A) \tkzGetSecondPoint{C}
\tkzDrawPolygon(A,B,C)
\tkzDrawSegment(O,C)
\tkzDrawPoints(A,B,C,O)
\tkzDrawSemiCircle(O,B)
\tkzLabelAngles[pos=1](B,A,C A,C,O){$\alpha$}
\tkzLabelAngles[pos=1.6](C,B,O O,C,B){$\beta$}
\tkzMarkAngles[mark=|,size=1.5](B,A,C A,C,O)
\tkzMarkAngles[mark=||,size=2](C,B,O O,C,B)
\tkzLabelPoints(A,B,O)
\tkzLabelPoints[above](C)
\end{tikzpicture}
\end{document}
答案3
使用纯净的tikz
:
\documentclass[tikz, border=3.141592]{standalone}
\usetikzlibrary{angles, arrows.meta,
intersections,
positioning,
quotes}
\usepackage{siunitx}
\begin{document}
\begin{tikzpicture}[
> = {Straight Barb[scale=0.8]},
myangle/.style = {angle radius=11mm,
angle eccentricity=0.8,
draw=#1!70!black, <->,
text=#1!70!black},
]
% triangle's coordinates
\coordinate (a);
\coordinate[right=6 of a] (b);
\coordinate[right=6 of b] (d);
% triangle's coordinate determined by intersection
\path[name path=ac] (a) -- ++ ( 60:6.5);
\path[name path=bc] (b) -- ++ (120:8);
\path[name intersections={of = ac and bc, by=c}];
% triangle's edges
\draw[semithick] (a) -- (c) -- (b) -- cycle
(b) -- (d) -- (c);
\draw (b) + (0,+1mm) -- +(0,-2mm) node[below] {S};
% circle
\draw (b) circle[radius=6];
% angles at A, B, C
\pic [myangle=teal, "$\alpha$"] {angle = b--a--c};
\pic [myangle=teal, "$\alpha$"] {angle = a--c--b};
% angles at C, D
\pic [myangle=cyan, "$\beta$"] {angle = c--d--b};
\pic [myangle=cyan, "$\beta$"] {angle = b--c--d};
\end{tikzpicture}
\end{document}
笔记:从几何学上可知,角度\alpha
可以是零度至九十度之间的任意角度。