有人能帮我在下一个文档的所有三行中生成三角形吗?我使用 \multirow 的所有尝试都失败了。
\documentclass{article}
\usepackage[utf8]{inputenc} % accents
\usepackage{verbatim} % \comment
\usepackage{amsmath,amssymb}
\usepackage{bm} % bold math
\usepackage{multirow,tabularx}
\usepackage{array} % \newcolumntype
\newcolumntype{x}[1]{ >{} m{#1} <{} }
\newcolumntype{X}[1]{ >{\[} m{#1} <{\]} }
%\newcolumntype{A}{>{$\begin{aligned}}c{\end{aligned}$}}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}
\tikzstyle{ciangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=1cm,
]
\tikzstyle{sqangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=1cm,
]
\newcommand{\DrawTriangle}[4]{%
\begin{tikzpicture}
\coordinate (A) at (-1.5,-1);
\coordinate (C) at (1.5,-1);
\coordinate (B) at (1.5,1);
\draw (C) -- node[right] {#1} (B) -- node[above] {#3} (A) -- node[below] {#2} (C);
\pic [ciangle, "#4"] {angle=C--A--B};
% \pic [ciangle, "#5"] {angle=A--B--C};
\draw [sqangle](C) rectangle ++(-0.5,0.5);
\end{tikzpicture}
}
\begin{document}
\begin{tabular}{x{3cm} X{1cm}}
\DrawTriangle{$a$}{$b$}{$c$}{$\alpha$} & \textit{sen}(\alpha)=\frac{a}{c} \\
& \textit{cos}(\alpha)=\frac{b}{c} \\
& \textit{tan}(\alpha)=\frac{a}{b}=\frac{\textit{sin}(\alpha)}{\textit{cos}(\alpha)} \\
\end{tabular}
\end{document}
另一个例子是:
\documentclass{article}
\usepackage[utf8]{inputenc} % accents
\usepackage{verbatim} % \comment
\usepackage{amsmath,amssymb}
\usepackage{bm} % bold math
\usepackage{array} % \newcolumntype
\newcolumntype{x}[1]{ >{} m{#1} <{} }
\newcolumntype{X}[1]{ >{\[} m{#1} <{\]} }
\usepackage{tikz}
\usetikzlibrary{angles,quotes}
\tikzstyle{ciangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=0.75cm,
]
\tikzstyle{sqangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
]
\newcommand{\DrawTriangle}[4]{%
\begin{tikzpicture}[scale=0.5]
\coordinate (A) at (-1.5,-1);
\coordinate (C) at (1.5,-1);
\coordinate (B) at (1.5,1);
\draw [sqangle](C) rectangle ++(-0.5,0.5);
\draw (C) -- node[right] {#1} (B) -- node[above] {#3} (A) -- node[below] {#2} (C);
\pic [ciangle, "#4"] {angle=C--A--B};
% \pic [ciangle, "#5"] {angle=A--B--C};
\end{tikzpicture}
}
\begin{document}
\begin{tabular}{|x{4cm}|x{2cm}|x{3cm}|}
\hline
Conocemos la longitud de dos lados, buscamos la longitud del tercer lado. & \DrawTriangle{$a$}{$b$}{$c?$}{}{} & Teorema Pitagoras \newline $ c^2=a^2+b^2 $ \\
& \DrawTriangle{$a?$}{$b$}{$c$}{} & \\
& \DrawTriangle{$a$}{$b?$}{$c$}{} & \\
\hline
\end{tabular}
\end{document}
答案1
\documentclass{article}
\usepackage[utf8]{inputenc} % accents
\usepackage{verbatim} % \comment
\usepackage{amsmath,amssymb}
\usepackage{bm} % bold math
\usepackage{array} % \newcolumntype
\newcolumntype{x}[1]{ >{} m{#1} <{} }
%\newcolumntype{A}{>{$\begin{aligned}}c{\end{aligned}$}}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}
\tikzstyle{ciangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=1cm,
]
\tikzstyle{sqangle}=[
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=1cm,
]
\newcommand{\DrawTriangle}[4]{%
\begin{tikzpicture}
\coordinate (A) at (-1.5,-1);
\coordinate (C) at (1.5,-1);
\coordinate (B) at (1.5,1);
\draw (C) -- node[right] {#1} (B) -- node[above] {#3} (A) -- node[below] {#2} (C);
\pic [ciangle, "#4"] {angle=C--A--B};
% \pic [ciangle, "#5"] {angle=A--B--C};
\draw [sqangle](C) rectangle ++(-0.5,0.5);
\end{tikzpicture}
}
\begin{document}
\begin{tabular}{x{3cm} x{2cm}}
\DrawTriangle{$a$}{$b$}{$c$}{$\alpha$} &
$\textit{sen}(\alpha)=\frac{a}{c}$% why \textit?
$\textit{cos}(\alpha)=\frac{b}{c}$
$\textit{tan}(\alpha)=\frac{a}{b}=\frac{\textit{sin}(\alpha)}{\textit{cos}(\alpha)} $
\end{tabular}
\end{document}
第二列需要更宽才能显示不换行的棕褐色线,我将其设置为 2 厘米而不是 1 厘米,但我不知道您的实际文档中有多少空间,所以我没有将其设置得更宽。
您没有使用tabularx
但已经加载了包,请注意,如果您加载tabularx
但定义自己的X
列类型,tabularx
则如果不进行一些内部重新定义,它将无法工作。