\documentclass[11pt,b4j,dvipdfmx,landscape,twocolumn,papersize]{jsarticle}
% furikana.sty	http://xymtex.com/fujitas2/texlatex/index.html#tategumi
\usepackage{ascmac,bm,amssymb,furikana}
\usepackage[deluxe]{otf}
\usepackage{graphicx,color}
%\usepackage{qrcode}
%\usepackage[pdfpagemode=FullScreen,setpagesize=false]{hyperref}
%\usepackage[pdfstartview={XYZ null null 1.75},setpagesize=false]{hyperref}
\usepackage{hyperref,pxjahyper}
\usepackage{tikz}
\usetikzlibrary{calc,angles}
%\usetikzlibrary{intersections,patterns,backgrounds,positioning}

%\AtBeginDvi{\special{pdf:pagesize width 364mm height 257mm}}
\textwidth=330truemm \textheight=226truemm \columnseprule=0pt
\topmargin=-24truemm \oddsidemargin=-8truemm

\hypersetup{% Oct 1 2023
	pdftitle={数学授業プリント},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/math_ia/math_ia.htm}
	}

\newenvironment{nidan}{\begin{minipage}[t]{.225\textwidth}}{\end{minipage}}
%\def\nidangumi#1#2{\begin{nidan}{#1}\end{nidan}\quad%
%\begin{nidan}{#2}\end{nidan}}
\newcommand{\nidangumi}[2]{\begin{nidan}{#1}\end{nidan}\quad%
\begin{nidan}{#2}\end{nidan}}

% ■■■ 0より大きいときはdvipdfmxで印刷用PDFを作ることを想定（答えはプリント
% 下部に出力される）
%        0 のときはdvipdfmxでデジタル教材用PDFを作ることを想定（答えは問題の
% すぐ横に赤で出力され、後でOneNoteで黒く塗りつぶし）
\def\make_pdf_with_insatsu{1}

\everymath{\displaystyle}
\parindent=0zw \columnsep=3zw
\lineskiplimit=1ex \normallineskiplimit=\lineskiplimit 
\lineskip=\lineskiplimit \normallineskip=\lineskiplimit

\newcounter{toi} \newcounter{subtoi}[toi]
\newcounter{subsubtoi}[subtoi]

\def\toi{\refstepcounter{toi}\par          % 問
  \makebox[2zw][r]{\Large\ajKaku{\value{toi}}\hspace{0.5zw}}%
  \hangindent=2zw}

\def\subtoi{\refstepcounter{subtoi}\par    % 小問
  \makebox[3.4zw][r]{\ajKakko{\value{subtoi}}\hspace{1zw}}%
  \hangindent=3.4zw}

\def\subsubtoi{\refstepcounter{subsubtoi}\par    % 小小問
  \makebox[4.2zw][r]{\ajMaru{\value{subtoi}}\hspace{1zw}}%
  \hangindent=4.2zw}

\ifnum\make_pdf_with_insatsu>0
	\def\kotae#1{\relax}
\else
	\def\kotae#1{\hfill \textcolor{red}{\footnotesize\gtfamily\bfseries\(\boldsymbol{#1} \)}}
\fi

\newcommand{\Slash}[1]{\ooalign{\hfil\kern-0.2em／\hfil\crcr$#1$}}
\pagestyle{empty}

\makeatletter
\def\hooklen#1#2{\settowidth{\@tempdima}{\(#1\)}
%\advance\@tempdima by.3ex
% ↑ 数式モードで式の前後に入るスペースを制御したかったが、
%    難しいのでやめた。段々難解なコードになっているのでやめた方がよい？
\hbox to\@tempdima{\hfil \(#2\)\hfil}}
\newcommand{\myleqq}{\mathrel{\mathpalette\gl@align<}}
\newcommand{\mygeqq}{\mathrel{\mathpalette\gl@align>}}
\newcommand{\gl@align}[2]{\lower.6ex\vbox{\baselineskip\z@skip\lineskip\z@
	\ialign{$\m@th#1\hfil##\hfil$\crcr#2\crcr=\crcr}}}
\makeatother


% ■■ 変数 yurika を定義する	2021 Feb 19
\newif\ifyurika
% \yurikatrue は有理化された答えが表示（\yurikafalse は有理化されない答え）
%\yurikatrue
\yurikafalse

\begin{document}
%
%  ===========================以下本文===============================
%
\def\namae{%
{\Large\gtfamily\bfseries 数学I \quad 授業プリント \# 46}\hfill
年\hskip8ex 組\hskip8ex 号 \hskip5ex

\hfill 
\underline{氏名\rule[-2ex]{0cm}{6ex}\hskip30ex}}

\namae

\vskip-4ex

{\large\gtfamily\bfseries ■　\kana{\(\mathstrut{\boldsymbol{\sin}}\)}{サイン}，
\kana{\(\mathstrut{\boldsymbol{\cos}}\)}{コサイン}，
\kana{\(\mathstrut{\boldsymbol{\tan}}\)}{タンジェント}}


\begin{center}
\begin{minipage}[t]{0.45\textwidth}
\begin{shadebox}

$\bullet$ \(\sin\mbox{A}={縦 \over 斜め} \)
\hskip6zw
$\bullet$ \(\cos\mbox{A}={横 \over 斜め} \)
\hskip6zw
$\bullet$ \(\tan\mbox{A}={\,縦\, \over 横} \)


\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｓｉｎ
\coordinate [label=below:A] (A) at (0.5,0);
\coordinate (B) at (2.5,1.2);
\coordinate (C) at (2.5,0);
\draw (A)--(B)--(C)--cycle;
\draw(C) rectangle ++(-0.2,0.2);
\draw[->,red,line width=2pt,line join=round] (0,0) .. controls (1,0.3) and (2,1) .. (3,2) .. controls (3,0.5) and (3,-0.1) .. (2.4,0);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}
%
\hspace*{2.4zw}
%
\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｃｏｓ
\coordinate [label=above:A] (A) at (0.5,0);
\coordinate (B) at (2.5,1.2);
\coordinate (C) at (2.5,0);
\draw (A)--(B)--(C)--cycle;
\draw(C) rectangle ++(-0.2,0.2);
\draw[->,red,line width=2pt,line join=round] (3,1.5) ..controls (1,1.7) and (-0.2,0.5) .. (0,0)
	..controls (0.2,-0.5) and (2,-0.5) .. (3,-0.1);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}
%
\hspace*{2.4zw}
%
\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｔａｎ
\coordinate [label=above:A] (A) at (0.5,0);
\coordinate (B) at (2.5,1.2);
\coordinate (C) at (2.5,0);
\draw (A)--(B)--(C)--cycle;
\draw(C) rectangle ++(-0.2,0.2);
\draw[->,red,line width=2pt,line join=round] (0,-0.2) ..controls (1,-0.2) and (3.2,-0.5) .. (3.2,0.3)
	..controls (3.2,0.7) and (3.2,1.5) .. (3.2,2);
\draw[red,line width=2pt,line join=round] (3.6,-0.2) ..controls (3.1,0) and (3.2,0.2) .. (3.2,1.9);
\draw[red,line width=2pt] (2.8,1.5) -- (3.6,1.5);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}


\end{shadebox}
\end{minipage}
\end{center}


\begin{center}
\begin{minipage}[t]{0.3\textwidth}
\begin{shadebox}

$\bullet$　三平方の定理

%%%\vspace*{1ex}

\begin{center}

\( \mbox{{\Huge\gtfamily\bfseries 斜め}}^2=○^2+△^2\)

\end{center}

\vspace*{0.5ex}

\end{shadebox}
\end{minipage}
\end{center}


\medskip
\ifyurika
\else
\quad{\Large\gtfamily\bfseries ※\ 三角比のときは【分母の有理化】は不要です。}
\fi
\bigskip

\unitlength=1mm


\toi 次の直角三角形ABCで，\(\sin A，\cos A，\tan A\)の値を求めなさい。

\nidangumi{\subtoi
\begin{tikzpicture}[scale=0.6,baseline=(current bounding box.north)]
\footnotesize
\coordinate [label=left:A] (A) at (0,0);
\coordinate [label=right:C] (C) at (4,0);
\coordinate [label=right:B] (B) at (4,3);
\small
%\draw[dashed] (A) to[out=-30,in=210] node[solid,inner sep=2pt, fill=white]{$4$} (C);
%\draw[dashed] (A) to[out=70,in=190] node[inner sep=2pt,fill=white]{$5$} (B);
%\draw[dashed] (C) to[out=45,in=-45] node[inner sep=2pt,fill=white]{$3$} (B);
%
%■ TikZ覚書.pdfを読んで \draw[bend right,distance=0.7cm](A) to (B) の書き方は知っていたが、曲がり具合が気に入らなくて（三角形みたいになる）ずっと to[out=,in=] を使っていた。改めて pgfmanual.pdf を読んでいて bend right=角度 の書き方ができることに気づいて、こちらの方が便利そうなのでコードを変更することにした。 May 10 2023
\draw[bend left=30,dashed](A) to node[fill=white]{$5$} (B)
				to node[fill=white]{$3$} (C)
				to node[fill=white]{$4$} cycle;
\draw (A)--(B)--(C)--cycle;
\draw (3.5,0)--(3.5,0.5)--(4,0.5);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}\\\hfill\kotae{\sin A={3\over5},\ \cos A={4\over5},
	\ \tan A={3\over4}}
}
{\subtoi
\begin{tikzpicture}[rotate=113, scale=0.26, baseline=(current bounding box.north)]
\footnotesize
\coordinate [label=left:A] (A) at (5,12);
\coordinate [label=right:B] (B) at (0,0);
\coordinate [label=above:C] (C) at (5,0);
\small
\draw[bend right=30,dashed](A) to node[fill=white]{$13$} (B)
				to node[fill=white]{$5$} (C)
				to node[fill=white]{$12$} cycle;
\draw (A)--(B)--(C)--cycle;
\draw (4,0)--(4,1)--(5,1);
\pic[draw=black, angle radius=12pt]{angle=B--A--C};
\end{tikzpicture}\\\hfill\kotae{\sin A={5\over13},\ \cos A={12\over13},\ \tan A={5\over12}}
}

\vfill


\nidangumi{\subtoi
\begin{tikzpicture}[scale=0.6,baseline=(current bounding box.north)]
\footnotesize
\coordinate [label=left:A] (A) at (0,3);
\coordinate [label=right:B] (B) at (2,0);
\coordinate [label=left:C] (C) at (0,0);
\draw[bend left=30,dashed](A) to node[fill=white]{$\sqrt{13}$} (B)
				to node[fill=white]{$2$} (C)
				to node[fill=white]{$3$} cycle;
\draw (A)--(B)--(C)--cycle;
\draw (0,0.3)--(0.3,0.3)--(0.3,0);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}\\\hfill\kotae{
\ifyurika
	\sin A={2\sqrt{13}\over13},\ \cos A={3\sqrt{13}\over13},\ \tan A={2\over3}
\else
	\sin A={2\over\sqrt{13}},\ \cos A={3\over\sqrt{13}},\ \tan A={2\over3}
\fi}
}
{\subtoi
\begin{tikzpicture}[rotate=-70, scale=0.9,baseline=(current bounding box.north)]
\footnotesize
\coordinate [label=right:A] (A) at ($ (1,{2*sqrt(2)}) $);% (1,2.82);
\coordinate [label=below:C] (C) at (1,0);
\coordinate [label=left:B] (B) at (0,0);
\small
\draw[bend right=30,dashed](A) to node[fill=white]{$3$} (B)
				to node[fill=white]{$1$} (C)
				to node[fill=white]{$2\sqrt{2}$} cycle;
\draw (A)--(B)--(C)--cycle;
\draw (0.7,0)--(0.7,0.3)--(1,0.3);
\pic[draw=black, angle radius=12pt]{angle=B--A--C};
\end{tikzpicture}\\\hfill\kotae{
\ifyurika
	\sin A={1\over3},\ \cos A={2\sqrt{2}\over3},\ \tan A={\sqrt{2}\over4}
\else
	\sin A={1\over3},\ \cos A={2\sqrt{2}\over3},\ \tan A={1\over2\sqrt{2}}
\fi}
}

\vfill


\newpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\toi 次の直角三角形を用いて\( 30^\circ,~45^\circ,~60^\circ \) の
\( \sin,~ \cos,~ \tan \)の値を求めなさい。\\
{\gtfamily\bfseries ※ 注　\( \boldsymbol{30^\circ,~45^\circ,~60^\circ} \)は
三角比の表を使ってはいけない}

\vskip-1zh

\hspace*{15mm}
\begin{tikzpicture}[rotate=0,scale=1.2,baseline=(current bounding box.south)]
%\coordinate [label=above:{\footnotesize B}] (B) at (1.732,1);
\coordinate (B) at ($ ({sqrt(3)},1) $);% (1.732,1);
\coordinate (A) at (0,0);
\coordinate (C) at ($ ({sqrt(3)},0) $);% (1.732,0);
\draw[bend left=30,dashed](A) to node[fill=white]{$2$} (B)
				to node[fill=white]{$1$} (C)
				to node[fill=white]{$\sqrt{3}$} cycle;
\draw(A)--(B)--(C)--cycle
	(C) rectangle ++(-0.2,0.2);
\draw(8pt,0) arc [start angle=0, end angle=30, radius=8pt];
\draw(10pt,0) arc [start angle=0, end angle=30, radius=10pt];
\footnotesize 
\draw(20pt,5pt) node{$30^\circ$};
\draw(B) +(0,-5pt) arc [start angle=270, end angle=210, radius=5pt];
\draw(B) +(-5pt,-10pt) node{$60^\circ$};
\end{tikzpicture}
%
\hspace*{7zw}
%
\begin{tikzpicture}[rotate=0,scale=1.2,baseline=(current bounding box.south)]
%\coordinate [label=above:{\footnotesize B}] (B) at (1,1);
\coordinate (B) at (1,1);
\coordinate (A) at (0,0);
\coordinate (C) at (1,0);
\draw[bend left=45,dashed](A) to node[fill=white]{$\sqrt{2}$} (B)
				to node[fill=white]{$1$} (C)
				to node[fill=white]{$1$} cycle;
\draw(A)--(B)--(C)--cycle
	(C) rectangle ++(-0.2,0.2);
\draw(5pt,0) arc [start angle=0, end angle=45, radius=5pt];
\footnotesize
\draw(12pt,5pt) node{$45^\circ$};
\draw(B) +(0,-5pt) arc [start angle=270, end angle=225, radius=5pt];
\draw(B) +(-5pt,-13pt) node{$45^\circ$};
\end{tikzpicture}
%
\hspace*{7zw}
%
\begin{tikzpicture}[rotate=0,scale=1.2,baseline=(current bounding box.south)]
\coordinate (B) at ($ (1,{sqrt(3)}) $);
\coordinate (A) at (0,0);
\coordinate (C) at (1,0);
\draw[bend left=30,dashed](A) to node[fill=white]{$2$} (B)
				to node[fill=white]{$\sqrt{3}$} (C)
				to node[fill=white]{$1$} cycle;
\draw(A)--(B)--(C)--cycle
	(C) rectangle ++(-0.2,0.2);
\draw(5pt,0) arc [start angle=0, end angle=60, radius=5pt];
\footnotesize
\draw(12pt,5pt) node{$60^\circ$};
\draw(B) +(0,-8pt) arc [start angle=270, end angle=240, radius=8pt];
\draw(B) +(0,-10pt) arc [start angle=270, end angle=240, radius=10pt];
\draw(B) +(-5pt,-20pt) node{$30^\circ$};
\end{tikzpicture}


\def\sikaku{\raisebox{0mm}{\fbox{\rule[-4mm]{0mm}{10mm}\hskip 8mm}}}
\def\naraberu#1{\( \makebox[7ex][l]{\(#1 30^\circ\)}= \) \sikaku\hfill
	\( \makebox[7ex][l]{\(#1 45^\circ\)} = \) \sikaku\hfill
	\( \makebox[7ex][l]{\(#1 60^\circ\)} =\) \sikaku\hfill}

\hfill\naraberu{\sin}\mbox{}

\hfill\naraberu{\cos}\mbox{}

\hfill\naraberu{\tan}\mbox{}


\kotae{\sin 30^\circ={1\over2},\sin 45^\circ={1\over\sqrt{2}},\sin 60^\circ={\sqrt{3}\over2}\quad 
{\sqrt{3}\over2},{1\over\sqrt{2}},{1\over2} \quad 
{1\over\sqrt{3}},1,\sqrt{3} }


\vskip1zh


\toi 三角比の表を使って，次の値を調べなさい。\\
\ajKakko{1}\quad \(\sin 25^\circ\) \hfill
\ajKakko{2}\quad \(\cos 50^\circ\) \hfill
\ajKakko{3}\quad \(\cos 74^\circ\) \hfill
\ajKakko{4}\quad \(\sin 18^\circ\) \hfill\mbox{}
\\[2ex]
\ajKakko{5}\quad \(\cos 32^\circ\) \hfill
\ajKakko{6}\quad \(\cos \phantom{7}7^\circ\) \hfill
\ajKakko{7}\quad \(\sin 83^\circ\) \hfill
\ajKakko{8}\quad \(\sin 58^\circ\) \hfill\mbox{}\\[2ex]



\kotae{ \mbox{\gt\ajKakko{1}}~0.4226 \ \mbox{\gt\ajKakko{2}}~0.6428 \ 
\mbox{\gt\ajKakko{3}}~0.2756 \quad \mbox{\gt\ajKakko{4}}~0.3090 \ 
\mbox{\gt\ajKakko{5}}~0.8480 \quad \mbox{\gt\ajKakko{6}}~0.9925 \ 
\mbox{\gt\ajKakko{7}}~0.9925 \quad \mbox{\gt\ajKakko{8}}~0.8480 }


\toi 次の直角三角形ABCで，\(\sin A，\cos A，\tan A\)の値を求めなさい。

\subtoi
\begin{tikzpicture}[scale=0.6,baseline=(current bounding box.north)]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize C}] (C) at (2,0);
\coordinate [label=right:{\footnotesize B}] (B) at ($ (2,{sqrt(5)}) $);% (2,2.23);
\draw (A)--(B)--(C)--cycle;
%\draw (1.7,0)--(1.7,0.3)--(2,0.3);
\draw (C) ++(-0.3,0)-- ++(0,0.3) -- ++(0.3,0);
\draw[dashed] (A) to[out=-45,in=225] node [inner sep=2pt, fill=white]{\small $2$} (C);
\draw[dashed] (A) to[out=80,in=200] node [inner sep=2pt, fill=white]{\small $3$} (B);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}\kotae{\sin A={\sqrt{5}\over3}, \ \cos A={2\over3}, \ \tan A={\sqrt{5}\over2}}

\vfill


\ifnum\make_pdf_with_insatsu=0
	\newpage
\fi


\subtoi
\begin{tikzpicture}[rotate=-60, scale=0.6,baseline=(current bounding box.north)]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=below:{\footnotesize C}] (C) at ($ ({sqrt(7)},0) $);% (2.65,0);
\coordinate [label=right:{\footnotesize B}] (B) at ($ ({sqrt(7)},3) $);% (2.65,3);
\draw (A)--(B)--(C)--cycle;
%\draw (2.35,0)--(2.35,0.3)--(2.65,0.3);
\draw (C) ++(-0.3,0)-- ++(0,0.3) -- ++(0.3,0);
\draw[dashed] (A) to[out=80,in=200] node [inner sep=2pt, fill=white]{\small $4$} (B);
\draw[dashed] (C) to[out=60,in=-60] node [inner sep=2pt, fill=white]{\small $3$} (B);
\pic[draw=black, angle radius=12pt]{angle=C--A--B};
\end{tikzpicture}\kotae{
\ifyurika
	\sin A={3\over4}, \ \cos A={\sqrt{7}\over4}, \ \tan A={3\sqrt{7}\over7}
\else
	\sin A={3\over4}, \ \cos A={\sqrt{7}\over4}, \ \tan A={3\over\sqrt{7}}
\fi}



\vfill






\subtoi
\begin{tikzpicture}[rotate=-60, scale=1, baseline=(current bounding box.north)]
\coordinate [label=right:{\footnotesize A}] (A) at (1,2);
\coordinate [label=below:{\footnotesize C}] (C) at (1,0);
\coordinate [label=left:{\footnotesize B}] (B) at (0,0);
\draw (A)--(B)--(C)--cycle;
%\draw (0.8,0)--(0.8,0.2)--(1,0.2);
\draw (C) ++(-0.2,0)-- ++(0,0.2) -- ++(0.2,0);
\draw[dashed] (B) to[out=-45,in=225] node [inner sep=2pt, fill=white]{\small $1$} (C);
\draw[dashed] (C) to[out=60,in=-60] node [inner sep=2pt, fill=white]{\small $2$} (A);
\pic[draw=black, angle radius=12pt]{angle=B--A--C};
\end{tikzpicture}\kotae{
\ifyurika
	\sin A={\sqrt{5}\over5}, \ \cos A={2\sqrt{5}\over5}, \ \tan A={1\over2}
\else
	\sin A={1\over\sqrt{5}}, \ \cos A={2\over\sqrt{5}}, \ \tan A={1\over2}
\fi}


\vfill

\newpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{tikzpicture}[scale=0.96]
\large
\filldraw[fill=blue, opacity=0.1] (0,1.5) rectangle (1,21.9)
			(7.6,1.5) rectangle (8.6,21.9);

\filldraw[fill=blue, opacity=0.2] (0,21.9) rectangle  (15.2,22.6);

\filldraw[fill=blue, opacity=0.8] (0,22.6) rectangle  (15.2,23.4);

\foreach \x in {0,1,3.2,5.4,7.6,8.6,10.8,13,15.2} {\draw(\x,1.5)--(\x,22.6);}
\foreach \y in {1.5,21.9,22.6,23.4} {\draw(0,\y)--(15.2,\y);}

\draw(0.5,22.25) node{$A$};\draw(2.1,22.25) node{$\sin A$};\draw(4.3,22.25) node{$\cos A$};\draw(6.5,22.25) node{$\tan A$};\draw(8.1,22.25) node{$A$};\draw(9.7,22.25) node{$\sin A$};\draw(11.9,22.25) node{$\cos A$};\draw(14.1,22.25) node{$\tan A$};

\draw(7.6,23) node{\textcolor{white}{\gt\Large 三角比の表}};

%%%%%%%%%%%%%%%%%%%%%%%%%　三角比の表　%%%%%%%%%%%%%%%%%%%%%%%%%
%\draw(0.5,21.6) node{$0^\circ$};\draw(2.1,21.6) node{$0.0000$};\draw(4.3,21.6) node{$1.0000$};\draw(6.5,21.6) node{$0.0000$};\draw(8.1,21.6) node{$45^\circ$};\draw(9.7,21.6) node{$0.7071$};\draw(11.9,21.6) node{$0.7071$};\draw(14.1,21.6) node{$1.0000$};
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\draw(0.5,21) node{$1^\circ$};\draw(2.1,21) node{$0.0175$};\draw(4.3,21) node{$0.9998$};\draw(6.5,21) node{$0.0175$};\draw(8.1,21) node{$46^\circ$};\draw(9.7,21) node{$0.7193$};\draw(11.9,21) node{$0.6947$};\draw(14.1,21) node{$1.0355$};
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\end{tikzpicture}




\ifnum\make_pdf_with_insatsu>0
\newpage
\mbox{}
\vfill

	\rotatebox{180}{\scriptsize
%	{\normalsize\ajKaku{1}}~\ajKakko{1}~\( {3\over5},{4\over5},{3\over4} \)\ 
%	\ajKakko{2}~\( {5\over13},{12\over13},{5\over12} \)\ 
%\ifyurika
%	\ajKakko{3}~\( {2\sqrt{13}\over13},{3\sqrt{13}\over13},{2\over3} \)\ 
%	\ajKakko{4}~\( {1\over3},{2\sqrt{2}\over3},{\sqrt{2}\over4} \)\ 
%\else
%	\ajKakko{3}~\( {2\over\sqrt{13}},{3\over\sqrt{13}},{2\over3} \)\ 
%	\ajKakko{4}~\( {1\over3},{2\sqrt{2}\over3},{1\over2\sqrt{2}} \)\ 
%\fi
%	{\normalsize\ajKaku{2}}~\( {1\over2},{1\over\sqrt{2}},{\sqrt{3}\over2}\ 
%	{\sqrt{3}\over2},{1\over\sqrt{2}},{1\over2} \ 
%	{1\over\sqrt{3}},1,\sqrt{3} \)\ 
	{\normalsize\ajKaku{3}}~\( 0.4226,~0.6428,~0.2756,~0.3090,~0.8480,
	~0.9925,~0.9925,~0.8480 \)\ 
	{\normalsize\ajKaku{4}}~\ajKakko{1}~\( {\sqrt{5}\over3},{2\over3},{\sqrt{5}\over2} \)\ 
\ifyurika
	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\sqrt{7}\over7} \)\ 
	\ajKakko{3}~\( {\sqrt{5}\over5},{2\sqrt{5}\over5},{1\over2} \)
\else
	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\over\sqrt{7}} \)\ 
	\ajKakko{3}~\( {1\over\sqrt{5}},{2\over\sqrt{5}},{1\over2} \)
\fi
	}

	\rotatebox{180}{{\normalsize\gtfamily\bfseries プリント\#46}\scriptsize
	{\normalsize\ajKaku{1}}~\ajKakko{1}~\( {3\over5},{4\over5},{3\over4} \)\ 
	\ajKakko{2}~\( {5\over13},{12\over13},{5\over12} \)\ 
\ifyurika
	\ajKakko{3}~\( {2\sqrt{13}\over13},{3\sqrt{13}\over13},{2\over3} \)\ 
	\ajKakko{4}~\( {1\over3},{2\sqrt{2}\over3},{\sqrt{2}\over4} \)\ 
\else
	\ajKakko{3}~\( {2\over\sqrt{13}},{3\over\sqrt{13}},{2\over3} \)\ 
	\ajKakko{4}~\( {1\over3},{2\sqrt{2}\over3},{1\over2\sqrt{2}} \)\ 
\fi
	{\normalsize\ajKaku{2}}~\( {1\over2},{1\over\sqrt{2}},{\sqrt{3}\over2}\ 
	{\sqrt{3}\over2},{1\over\sqrt{2}},{1\over2} \ 
	{1\over\sqrt{3}},1,\sqrt{3} \)\ 
%	{\normalsize\ajKaku{3}}~\( 0.4226,~0.6428,~0.2756,~0.3090,~0.8480,
%	~0.9925,~0.9925,~0.8480 \)\ 
%	{\normalsize\ajKaku{4}}~\ajKakko{1}~\( {\sqrt{5}\over3},{2\over3},{\sqrt{5}\over2} \)\ 
%\ifyurika
%	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\sqrt{7}\over7} \)\ 
%	\ajKakko{3}~\( {\sqrt{5}\over5},{2\sqrt{5}\over5},{1\over2} \)
%\else
%	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\over\sqrt{7}} \)\ 
%	\ajKakko{3}~\( {1\over\sqrt{5}},{2\over\sqrt{5}},{1\over2} \)
%\fi
	}
\fi





\end{document}
