%
%	gbb60166@gmail.com	https://unilab.gbb60166.jp/prekou/prekou.htm
%
% aspectratio= は 1610, 169, 149, 54, 32 の中から選べる（省略時は 43）
% C:\texlive\2024\texmf-dist\tex\latex\beamer\beamer.cls
%\documentclass[20pt,dvipdfmx,usepdftitle=false,hyperref={pdfpagemode=FullScreen},aspectratio=169]{beamer}
%\documentclass[20pt,dvipdfmx,usepdftitle=false,hyperref={pdfstartview={XYZ null null 8.00}},aspectratio=169]{beamer}
\documentclass[20pt,dvipdfmx,usepdftitle=false,aspectratio=169]{beamer}
% pdfの栞の字化けを防ぐ
%\AtBeginDvi{\special{pdf:tounicode EUC-UCS2}}
% テーマ
\usetheme{Copenhagen}
% navi. symbolsは目立たないが，dvipdfmxを使うと機能しないので非表示に
\setbeamertemplate{navigation symbols}{} 
\usepackage{bxdpx-beamer,pxjahyper,minijs}
%\usepackage{graphicx}
%\usepackage{tkokugo,furikana,tsayusen,shiika,sfkanbun,jdkintou,plext}
\usepackage{type1cm}
\usepackage[deluxe]{otf}
%\usepackage{tikzsymbols}
%\usepackage[dvipdfmx]{graphicx}%
\def\pgfsysdriver{pgfsys-dvipdfmx.def}%（graphicxパッケージを使用しない場合はこの行を有効に）
%\def\pgfsysdriver{pgfsys-dvips.def}%デフォルト
%\usepackage[e]{esvect}
\usepackage{tikz}%（これで、pgfとpgfforが読み込まれます。）
%\usepackage{tikz-3dplot}
\usetikzlibrary{calc}
%\usetikzlibrary{arrows.meta,patterns,angles,intersections,backgrounds}
%\usepackage{animate}
%\usepackage{pgfplots}
%\pgfplotsset{compat=newest}
%\pgfplotsset{compat=newest, every axis/.append style={line width=1pt}}
%\PassOptionsToPackage{dvipdfmx}{graphicx}
\nofiles

\hypersetup{% Jul 18 2023
	pdftitle={プレ高数学科（LaTeX Beamer で汎用PDFプレゼンを作る）},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/prekou/prekou.htm}
	}

% フォントはお好みで
%\usepackage{txfonts}

\mathversion{bold}
\renewcommand{\familydefault}{\sfdefault}
\renewcommand{\seriesdefault}{bx}
\renewcommand{\kanjifamilydefault}{\gtdefault}
\renewcommand{\kanjiseriesdefault}{bx}
\setbeamerfont{title}{size=\normalsize,series=\bfseries}
\setbeamerfont{frametitle}{size=\normalsize,series=\bfseries}
\setbeamertemplate{frametitle}[default][center]
\usefonttheme{professionalfonts}

% 参考にしたURL
% http://windom.phys.hirosaki-u.ac.jp/fswiki/wiki.cgi?page=LaTeX+Beamer%A4%C7%A5%D7%A5%EC%A5%BC%A5%F3%A5%C6%A1%BC%A5%B7%A5%E7%A5%F3

\newcommand{\Slash}[1]{\ooalign{\hfil\kern-3pt／\hfil\crcr$#1$}}
\everymath{\displaystyle}

\newcommand{\maruwaku}[1]{\begin{tikzpicture}[scale=0.7, baseline={([yshift=-22pt] current bounding box.north)}]
\filldraw[color=CUDBlue, line width=1pt, rounded corners=2pt] (-0.1,0)--(2.1,0)--(2.1,1.1)--(-0.1,1.1)--cycle;
\draw(1,0.5) node[white]{#1};
\end{tikzpicture}
}

\setbeamersize{text margin left=5mm,text margin right=5mm}
\fboxrule=0.8pt

\makeatletter
\newcommand{\hooklen}[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}}
%■等号付き不等号 by 美文書作成入門
\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

% カラーユニバーサルデザインを調べたつもりだがあまり自信がありません
% http://www.fukushihoken.metro.tokyo.jp/kiban/machizukuri/kanren/color.files/colorudguideline.pdf
% http://jfly.iam.u-tokyo.ac.jp/colorset/

% ■ アクセントカラー小面積を目立たせる高彩度色
\definecolor{CUDRed}{RGB}{255,75,0}
\definecolor{CUDGreen}{RGB}{3,175,122}
\definecolor{CUDBlue}{RGB}{0,90,255}
\definecolor{CUDCyan}{RGB}{77,196,255}
\definecolor{CUDMagenta}{RGB}{153,0,153}
\definecolor{CUDYellow}{RGB}{255,241,0}
\definecolor{CUDBrown}{RGB}{128,64,0}

\definecolor{CUDOrange}{RGB}{246,170,0}

% ■ ベースカラー広い面積の塗り分けに用いる低・中彩度色
\definecolor{CUDPink}{RGB}{255,202,191}
\definecolor{CUDBrightGreen}{RGB}{119,217,168}
\definecolor{CUDLime}{RGB}{216,242,85}
\definecolor{CUDCream}{RGB}{255,255,128}
\definecolor{CUDBrightCyan}{RGB}{191,228,255}
%\definecolor{CUD}{RGB}{}%

\setbeamercolor{CUDBrightGreen}{fg=black,bg=CUDBrightGreen!50}
\setbeamercolor{CUDBrightCyan}{fg=black,bg=CUDBrightCyan}
\setbeamercolor{CUDCream}{fg=black,bg=CUDCream!75}

%\setbeamercolor{frametitle}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{palette primary}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{frametitle}{bg=CUDOrange}
%\setbeamercolor{palette primary}{bg=CUDOrange}
%\setbeamercolor{frametitle}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{palette primary}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{frametitle}{bg=CUDCream!75,fg=black!80}
%\setbeamercolor{palette primary}{bg=CUDCream!75,fg=black!80}

\newcommand{\myfrac}[2]{\mathchoice%■ 修正 Oct 8,2023
	{%\fontsize{20}{20}\selectfont%	\displaystyle
		\frac{\raisebox{-.6ex}{\(#1\)}}{\raisebox{.4ex}{\(#2\)}}}%
	{\textstyle{\frac{#1}{#2}}}%	\textstyle
	{%\fontsize{13}{13}\selectfont%	\scriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	{%\fontsize{11}{11}\selectfont%	\scriptscriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	}


\begin{document}

\title{プレ高数学科}\author{gbb60166}

%■ 本文と数式モードの間に挿入される空白スペース
%\mathsurround=-1pt
%■ \( \log\log \)でlogとlogの間に挿入される細スペース（省略形は \,）
%\thinmuskip=3mu
%■ \( a+b \)で+の前後に挿入される中スペース（省略形は \>）
%\medmuskip=2mu plus 1mu minus 1mu
\medmuskip=2mu minus 1mu
%■ \( x=5 \)で=の前後に挿入される太スペース（省略形は \;）
%\thickmuskip=5mu plus 5mu
% mu は math units の省略で長さの単位

%■ 新しいプラスマイナス記号を自作（TikZを利用）
\newcommand{\newpm}{\,\tikz[baseline=-0.1ex]
	\draw[line width=0.1ex,line cap=round](0,0)--++(0.6em,0)
					(0,0.8ex)--++(0.6em,0)
					(0.3em,0.25ex)--++(0,1.1ex);
	\,}
\newcommand{\henao}{\tikz[baseline=-6pt]\draw(0,0)
	node[white,circle, fill=CUDBlue,inner sep=0pt]
		{\small\gtfamily\bfseries 辺};}

%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■


%\end{document}
%■■■■■■■■■■■■■　完成品　■■■■■■■■■■■■■■

\begin{frame}[t]
\frametitle{余弦定理}
\vspace*{-1.5zh}
\[ a^2=b^2+c^2-2bc\cos A \]
\[ b^2=c^2+a^2-2ca\cos B \]
\[ c^2=a^2+b^2-2ab\cos C \]

\bigskip
\centerline{\textcolor{CUDRed}{こんな公式は覚える必要なし！}}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{余弦定理}
\vspace*{-1.5zh}
\[a^2=b^2+c^2-2bc\cos A \]

だけ覚えて\ 
\begin{tikzpicture}[CUDRed,baseline=-5pt,thick,scale=2,>=stealth,very thick]
\draw[->](60:10pt) arc (60:-10:10pt);
\draw[->](300:10pt) arc (300:230:10pt);
\draw[->](180:10pt) arc (180:110:10pt);
\draw(0,10pt) node{$a$};
\draw(-30:10pt) node[CUDRed]{$b$};
\draw(210:10pt) node[CUDRed]{$c$};
\end{tikzpicture}
と置き換えろ！とする場合もあるが、それでは
\textcolor{CUDBlue}{この公式の本質を理解しているとは思えない。}

\bigskip
\centerline{\textcolor{CUDRed}{数学は暗記科目じゃないぞ！考える学問だ！}}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{余弦定理}
%\vskip-2zh
\( \text{\textcolor{CUDGreen}{角度の向かいの辺}}
	^{\raisebox{.2ex}{\(\scriptstyle2\)}} \)\\[1ex]

\hfill\( =\ \colorbox{CUDPink}{\( \henao^2  +  \henao^2 \)}
		-2\times\colorbox{CUDCream}{%
		\(\henao  \times  \henao \)}  \times 
			\cos\text{\textcolor{CUDRed}{間の角度}} \)

%\begin{eqnarray*}
%\lefteqn{\text{\textcolor{CUDGreen}{角度の向かいの辺}}^2}\\[1zh]
%	&=& \colorbox{CUDPink}{\( \henao^2  +  \henao^2 \)}
%		-2\times\colorbox{CUDCream}{%
%		\(\henao  \times  \henao \)}  \times 
%			\cos\text{\textcolor{CUDRed}{間の角度}}
%\end{eqnarray*}

\vskip4ex
\centerline{\textcolor{CUDRed}{覚えるのはこれひとつ}}
\centerline{\textcolor{CUDRed}{こっちがわかりやすい（たぶん…）}}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい（例題\(1\)）}
\vskip-4ex
\begin{tikzpicture}[scale=1,thick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](0,0) grid (15,7);
\useasboundingbox(0,0) grid (15,7);

\fill[CUDPink](3.8,5.5) rectangle ++(3.2,1);
\fill[CUDCream](8.8,5.5) rectangle ++(2.6,1);

\draw(7.5,6) node{\( {\kern1em}^2={\kern1em}^2 \!+{\kern1em}^2
		-2\times\kern1em \times\kern1em \times\cos \)};

\alt<2->{\draw[CUDGreen](2.3,6) node{\( x \)};
	\draw[CUDBlue](4.5,6) node{\( 2 \)}
		(6.3,6) node{\( 3 \)}
		(9.3,6) node{\( 2 \)}
		(10.9,6) node{\( 3 \)}
		(11,3) node[text width=14zw,align=center]
			{\(2\)と\(3\)の順不同\\[-.5ex]
			\((3\)と\(2\)でもOK\()\)};
	\draw[CUDRed](13.9,6) node{\( 60^\circ \)};
	}
	{\draw[CUDGreen](1.2,6) node{\small 向かいの辺};
	\draw(4.3,6) node{\henao}
		(6.2,6) node{\henao}
		(9.3,6) node{\henao}
		(10.9,6) node{\henao};
	\draw[CUDRed](13.9,6) node{角度};
	}
\begin{scope}[thick,scale=2,xshift=0.3em,yshift=2ex,line join=round,
	gbb/.style={fill=white,inner sep=1pt}]
\coordinate (O) at (0,0);
\coordinate (A) at (60:2);%極座標（角度：半径）
\coordinate (B) at (3,0);
%\normalsize

\draw[dashed] (O) to[out=-30,in=210] node[CUDBlue,gbb]{\(3\)} (B);
\draw[dashed] (O) to[out=90,in=210] node[CUDBlue,gbb]{\(2\)} (A);
\draw[dashed] (A) to[out=0,in=100] node[CUDGreen,gbb]{\(x\)} (B);

\draw[CUDRed,line width=2pt](10pt,0) 
	arc [start angle=0, end angle=60,radius=10pt];
\draw(0.7,0.3) node[CUDRed]{\(60^\circ\)};

\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
\draw[line width=2pt,line cap=round](A)--(B);
\end{scope}

\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\begin{frame}[t]
%\frametitle{辺\(x\)の長さを求めなさい（例題$1$）}
%\vspace*{-2zh}
%\only<1>{\vspace*{3pt}\small
%	\[ \text{\textcolor{CUDGreen}{向かい辺}}^2= %
%	\colorbox{CUDPink}{\( \henao^2  + \henao^2 \)}
%			-2\times\colorbox{CUDCream}{\(
%			\henao  \times  \henao \)}  \times 
%				\cos\text{\textcolor{CUDRed}{角度}} \]
%	}
%
%\only<2-3>{\[ %%\vphantom{\colorbox{CUDPink}{\( \henao^2  + \henao^2 \)}}
%	\makebox[4zw]{\( \textcolor{CUDGreen}{x}^2 \)} =
%	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{2}^2
%		+\textcolor{CUDBlue}{3}^2 \)}
%	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{2}\times
%	\textcolor{CUDBlue}{3} \)}\times\cos\textcolor{CUDRed}{60^\circ} \]}
%
%\only<4>{\[ \makebox[4zw]{\( \textcolor{CUDGreen}{x}^2 \)} =
%	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{3}^2
%		+\textcolor{CUDBlue}{2}^2 \)}
%	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{3}\times
%	\textcolor{CUDBlue}{2} \)}\times\cos\textcolor{CUDRed}{60^\circ} \]}
%
%\only<5>{\[ \makebox[4zw]{\( \textcolor{CUDGreen}{x}^2 \)} =
%	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{3}^2
%		+\textcolor{CUDBlue}{2}^2 \)}
%	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{2}\times
%	\textcolor{CUDBlue}{3} \)}\times\cos\textcolor{CUDRed}{60^\circ} \]}
%
%\begin{minipage}{0.45\textwidth}
%\begin{tikzpicture}[thick,scale=2,line join=round,
%	gbb/.style={fill=white,inner sep=1pt}]
%\coordinate (O) at (0,0);
%\coordinate (A) at (60:2);%極座標（角度：半径）
%\coordinate (B) at (3,0);
%\normalsize
%
%\draw[dashed] (O) to[out=-30,in=210] 
%	node[CUDBlue,gbb]{\(3\)} (B);
%\draw[dashed] (O) to[out=90,in=210] 
%	node[CUDBlue,gbb]{\(2\)} (A);
%\draw[dashed] (A) to[out=0,in=100] 
%	node[CUDGreen,gbb]{\(x\)} (B);
%
%\draw[CUDRed,line width=2pt](10pt,0) 
%	arc [start angle=0, end angle=60,radius=10pt];
%\draw(0.7,0.3) node[CUDRed]{\(60^\circ\)};
%
%\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
%\draw[line width=2pt,line cap=round](A)--(B);
%
%\visible<3->{\draw[CUDBlue](4.8,1.5) node{$2$と$3$どちらが先でもOK}
%			(5,1) node{$( \ 2$と$3$がペアが重要 $)$};
%	}
%\end{tikzpicture}
%\end{minipage}
%\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい\qquad
\visible<2->{%
\begin{tikzpicture}[scale=0.8, thick,
	gbb/.style={fill=white,inner sep=1pt},
	baseline={([yshift=-3.5ex] current bounding box.north)}]
%\fill[CUDCream](-1,-0.3) rectangle (1.8,1.8);
\draw[text=CUDRed,rounded corners=.3pt](0,0)-- 
	node[pos=0.6,gbb]{\tiny$1$}
	(1,0)-- node[xshift=6pt,gbb]{\tiny$\sqrt{3}$}
	(1,{sqrt(3)})-- node[gbb,xshift=-1ex]{\tiny$2$}
				cycle;
\draw(10pt,0) arc [start angle=0, end angle=60,radius=10pt];
\draw(-10pt,6pt) node{\tiny\( 60^\circ \)};
\end{tikzpicture}
}}
\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& 2^2+3^2
\tikz[baseline=-0.3ex,very thick]{%
		\useasboundingbox(0,0);%縦0横0の描画領域
		\draw[CUDOrange,dashed](-3em,-13ex) rectangle ++(3em,15ex);
		\draw[CUDMagenta,double,thin](1em,-13ex)
			rectangle ++(8em,15ex);
		}
	-2\times2\times3\times\cos60^\circ \\\pause\pause
x^2 &=& \hooklen{2^2}{4}+\hooklen{3^2}{9} 
	-2\times2\times3\times\hooklen{cos60^\circ}{\myfrac{\,1\,}{2}} \\
								\pause
x^2 &=& \hooklen{2^2+3^2}{13} 
	-2\hskip-16pt\tikz \draw[line width=2.4pt,CUDRed](0pt,0pt) 
		--(16pt,14pt);
	\times2\times3\times\hooklen{cos60^\circ}
		{\myfrac{\,1\,}{2\hskip-16pt\tikz
	\draw[line width=2.4pt,CUDRed] (0pt,0pt) -- (16pt,14pt);}} \\ \pause
x^2 &=& \hooklen{2^2+3^2}{13} - \hooklen{2\times2\times3\times0}{6} \\ \pause
x^2 &=& \hooklen{2^2+3^2 -2\times2\times3}{7}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい}
\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& 7 \\\pause
x &=& \newpm\sqrt{7}
\end{eqnarray*}
\qquad \(x>0\)なので \pause \hspace*{4ex}\(x=\sqrt{7}\)\quad\□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい（例題\(2\)）}
\vskip-4ex
\begin{tikzpicture}[scale=1,thick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](0,0) grid (15,7);
\useasboundingbox(0,0) grid (15,7);

\fill[CUDPink](3.8,5.5) rectangle ++(3.4,1);
\fill[CUDCream](9.1,5.5) rectangle ++(2.8,1);

\draw(7.7,6) node{\( {\kern1em}^2={\kern1.3em}^2 \!+{\kern1em}^2
		-2\times\kern1.3em \times\kern1em \times\cos \)};

\alt<2->{\draw[CUDGreen](2.2,6) node{\( x \)};
	\draw[CUDBlue](4.3,6.1) node{\( \sqrt{2} \)}
		(6.5,6) node{\( 3 \)}
		(9.6,6.1) node{\( \sqrt{2} \)}
		(11.3,6) node{\( 3 \)}
		(11,3) node[text width=14zw,align=center]
			{\(\sqrt{2}\)と\(3\)の順不同\\
			\((3\)と\(\sqrt{2}\)でもOK\()\)};
	\draw[CUDRed](14.2,6) node{\( 45^\circ \)};
	}
	{\draw[CUDGreen](1.2,6) node{\small 向かいの辺};
	\draw(4.5,6) node{\henao}
		(6.4,6) node{\henao}
		(9.7,6) node{\henao}
		(11.3,6) node{\henao};
	\draw[CUDRed](14.3,6) node{角度};
	}
\begin{scope}[thick,scale=2,xshift=4em,yshift=4ex,line join=round,
	gbb/.style={fill=white,inner sep=1pt}]
\coordinate (O) at (0,0);
\coordinate (A) at ($ (135:{sqrt(2)}) $);%極座標（角度：半径）
\coordinate (B) at (-3,0);
\normalsize

\draw[dashed] (O) to[out=210,in=-30] 
	node[CUDBlue,fill=white,inner sep=2pt]{\(3\)} (B);
\draw[dashed] (O) to[out=90,in=0] 
	node[CUDBlue,fill=white,inner sep=0pt]{\(\sqrt{2}\)} (A);
\draw[dashed] (A) to[out=180,in=60] 
	node[CUDGreen,fill=white,inner sep=1pt]{\(x\)} (B);

\draw[CUDRed,line width=2pt](-10pt,0) 
	arc [start angle=180, end angle=135,radius=10pt];
\draw(-0.7,0.2) node {\textcolor{CUDRed}{$45^\circ$}};

\draw[line width=2pt,line cap=round](A)--(B);
\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
\end{scope}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\begin{frame}[t]
%\frametitle{辺\(x\)の長さを求めなさい（例題$2$）}
%\vspace*{-2zh}
%\alt<2->{\[ \makebox[4zw]{\( \textcolor{CUDGreen}{x}^2 \)} =
%	\colorbox{CUDPink}{\( \mathstrut{
%	\textcolor{CUDBlue}{\sqrt{2}}}^{\kern4pt 2}
%		+\textcolor{CUDBlue}{3}^2 \)}
%	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{\sqrt{2}}\times
%	\textcolor{CUDBlue}{3} \)}\times\cos\textcolor{CUDRed}{45^\circ} \]}
%		{\vspace*{5pt}\small
%		\[ \text{\textcolor{CUDGreen}{向かい辺}}^2= %
%		\colorbox{CUDPink}{\( \henao^2  + \henao^2 \)}
%			-2\times\colorbox{CUDCream}{%
%			\( \henao  \times  \henao \)}  \times 
%				\cos\text{\textcolor{CUDRed}{角度}} \]
%\vspace*{-12.5pt}
%		}
%
%\begin{minipage}{0.45\textwidth}
%\begin{tikzpicture}[thick,scale=2, line join=round]
%\coordinate (O) at (0,0);
%\coordinate (A) at ($ (135:{sqrt(2)}) $);%極座標（角度：半径）
%\coordinate (B) at (-3,0);
%\normalsize
%
%\draw[dashed] (O) to[out=210,in=-30] 
%	node[CUDBlue,fill=white,inner sep=2pt]{\(3\)} (B);
%\draw[dashed] (O) to[out=90,in=0] 
%	node[CUDBlue,fill=white,inner sep=0pt]{\(\sqrt{2}\)} (A);
%\draw[dashed] (A) to[out=180,in=60] 
%	node[CUDGreen,fill=white,inner sep=1pt]{\(x\)} (B);
%
%\draw[CUDRed,line width=2pt](-10pt,0) 
%	arc [start angle=180, end angle=135,radius=10pt];
%\draw(-0.7,0.2) node {\textcolor{CUDRed}{$45^\circ$}};
%
%\draw[line width=2pt,line cap=round](A)--(B);
%\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
%
%\small 
%\visible<2->{\draw[CUDBlue](2.3,1) node{$\sqrt{2}$と$3$どちらが先でもOK}
%		(2.5,0.5) node{$( \ \sqrt{2}$と$3$がペアが重要\ $)$};
%	}
%\end{tikzpicture}
%\end{minipage}
%\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい\qquad
\visible<2->{%
\begin{tikzpicture}[scale=0.8, thick,
	gbb/.style={fill=white,inner sep=1pt},
	baseline={([yshift=-2ex] current bounding box.north)}]
%\fill[CUDCream](-1,-0.4) rectangle (5.2,1.5);
\draw[text=CUDRed,rounded corners=.3pt](0,0)-- 
	node[pos=0.6,gbb]{\tiny$1$}
	(1,0)-- node[gbb,xshift=6pt]{\tiny$1$}
	(1,1)-- node[gbb,xshift=-1ex,yshift=6pt]
		{\tiny$\sqrt{2}$}  cycle;
\draw(10pt,0) arc [start angle=0, end angle=45,radius=10pt];
\draw(-10pt,6pt) node{\tiny\( 45^\circ \)};
\end{tikzpicture}
}}
\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& \sqrt{2}^{\kern4pt 2}+3^2
\tikz[baseline=-0.3ex,very thick]{%
		\useasboundingbox(0,0);%縦0横0の描画領域
		\draw[CUDOrange,dashed](-4em,-13.5ex) rectangle ++(4em,16ex);
		\draw[CUDMagenta,double,thin](1em,-13.5ex)
				rectangle ++(8.6em,16ex);
		}
	-2\times\sqrt{2}\times3\times\cos45^\circ \\ \pause\pause
x^2 &=& \hooklen{\sqrt{2}^{\kern4pt 2}}{2}+\hooklen{3^2}{9} 
	-2\times\sqrt{2}\times3\times\hooklen{\cos45^\circ}
		{\myfrac{\,1\,}{\sqrt{2}}} \\  \pause
x^2 &=& \hooklen{\sqrt{2}^{\kern4pt 2}+3^2}{11} 
	-2\times{\sqrt{2}\hskip-32pt\tikz 
		\draw[line width=2.4pt,CUDRed] (0pt,0pt)--(36pt,18pt);}
	\times3\times\hooklen{\cos45^\circ}{\myfrac{\,1\,}
	{{\sqrt{2}\hskip-32pt\tikz \draw[line width=2.4pt,CUDRed](0pt,0pt)
		--(36pt,18pt);}}} \\  \pause
x^2 &=& \hooklen{\sqrt{2}^{\kern4pt 2}+3^2}{11} 
	- \hooklen{2\times\sqrt{2}\times3}{6} \\  \pause
x^2 &=& \hooklen{\sqrt{2}^2+3^2 -2\times\sqrt{2}\times3}{5}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい}

\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& 5 \\\pause
x &=& \newpm\sqrt{5}
\end{eqnarray*}

\qquad \(x>0\)なので \pause \hspace*{4ex}\(x=\sqrt{5}\)\quad\□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい（例題\(3\)）}
\vskip-4ex
\begin{tikzpicture}[scale=1,thick, >=stealth,line join=round,
	gbb/.style={fill=white,inner sep=2pt},
	baseline={([yshift=0ex] current bounding box.north)}]
%\draw[help lines](0,0) grid (15,7);
\useasboundingbox(0,0) grid (15,7);

\fill[CUDPink](3.8,5.5) rectangle ++(3.2,1);
\fill[CUDCream](8.8,5.5) rectangle ++(2.6,1);

\draw(7.5,6) node{\( {\kern1em}^2={\kern1em}^2 \!+{\kern1em}^2
		-2\times\kern1em \times\kern1em \times\cos \)};

\alt<2->{\draw[CUDGreen](2.3,6) node{\( x \)};
	\draw[CUDBlue](4.5,6) node{\( 5 \)}
		(6.3,6) node{\( 4 \)}
		(9.3,6) node{\( 5 \)}
		(10.9,6) node{\( 4 \)}
		(11,3) node[text width=14zw,align=center]
			{\(5\)と\(4\)の順不同\\[-.5ex]
			\((4\)と\(5\)でもOK\()\)};
	\draw[CUDRed](14.1,6) node{\( 120^\circ \)};
	}
	{\draw[CUDGreen](1.2,6) node{\small 向かいの辺};
	\draw(4.3,6) node{\henao}
		(6.2,6) node{\henao}
		(9.3,6) node{\henao}
		(10.9,6) node{\henao};
	\draw[CUDRed](13.9,6) node{角度};
	}
\begin{scope}[thick,scale=1,xshift=4em,yshift=13.5ex,line join=round,
	gbb/.style={fill=white,inner sep=3pt}]
\coordinate (O) at (0,0);
\coordinate (A) at (4,0);
\coordinate (B) at (-120:5);%極座標（角度：半径）
\normalsize

\draw[dashed] (O) to[out=30,in=150] node[CUDBlue,gbb]{\(4\)} (A);
\draw[dashed] (O) to[out=210,in=90] node[CUDBlue,gbb]{\(5\)} (B);
\draw[dashed] (B) to[out=10,in=250] node[CUDGreen,gbb]{\(x\)} (A);

\draw[CUDRed,line width=2pt](16pt,0)
	arc[start angle=0, end angle=-120, radius=16pt];
\draw(1,-1) node {\textcolor{CUDRed}{$120^\circ$}};

\draw[line width=2pt,line cap=round](A)--(B);
\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
\end{scope}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\begin{frame}[t]
%\frametitle{辺\(x\)の長さを求めなさい（例題$3$）}
%\vspace*{-2zh}
%\alt<2->{\[ \makebox[4zw]{\( \textcolor{CUDGreen}{x}^2 \)} =
%	\colorbox{CUDPink}{\( \mathstrut{\textcolor{CUDBlue}{5}}^2+
%		\textcolor{CUDBlue}{4}^2 \)}
%	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{5}\times
%	\textcolor{CUDBlue}{4} \)}\times\cos\textcolor{CUDRed}{120^\circ} \]}
%		{\vspace*{5pt}\small
%		\[ \text{\textcolor{CUDGreen}{向かい辺}}^2= %
%		\colorbox{CUDPink}{\( \henao^2  + \henao^2 \)}
%			-2\times\colorbox{CUDCream}{%
%			\( \henao  \times  \henao \)}  \times 
%				\cos\text{\textcolor{CUDRed}{角度}} \]
%\vspace*{-18pt}
%		}
%
%\begin{minipage}{0.45\textwidth}
%\begin{tikzpicture}[thick,scale=1, line join=round,
%	gbb/.style={fill=white,inner sep=3pt}]
%\coordinate (O) at (0,0);
%\coordinate (A) at (4,0);
%\coordinate (B) at (-120:5);%極座標（角度：半径）
%\normalsize
%
%\draw[dashed] (O) to[out=30,in=150]
%	node[CUDBlue,gbb]{\(4\)} (A);
%\draw[dashed] (O) to[out=210,in=90]
%	node[CUDBlue,gbb]{\(5\)} (B);
%\draw[dashed] (B) to[out=10,in=250]
%	node[CUDGreen,gbb]{\(x\)} (A);
%
%\draw[CUDRed,line width=2pt](16pt,0)
%	arc[start angle=0, end angle=-120, radius=16pt];
%\draw(1,-1) node {\textcolor{CUDRed}{$120^\circ$}};
%
%\draw[line width=2pt,line cap=round](A)--(B);
%\draw[CUDBlue,line width=2pt,line cap=round](A)--(O)--(B);
%\end{tikzpicture}
%\end{minipage}
%\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい\qquad
\visible<2->{%
\begin{tikzpicture}[scale=0.8, thick, >=stealth,
	gbb/.style={fill=white,inner sep=1pt},
	baseline={([yshift=-3.5ex] current bounding box.north)}]
\tiny
\draw[->](-1.5,0)--++(2.4,0);
\draw[->](0,-0.2)--++(0,{sqrt(3)});
\draw[text=CUDRed](0,0)
	-- node[xshift=2pt,gbb]{\(2\)}(-1,{sqrt(3)})
	-- node[xshift=-10pt,fill=white,inner sep=0pt]{\(\sqrt{3}\)} (-1,0);
\draw[CUDRed](-0.64,0.06) node[xshift=2pt,gbb]{\(-1\)};
\draw(10pt,0) arc [start angle=0, end angle=120,radius=10pt];
\draw(20pt,16pt) node{\( 120^\circ \)};
\end{tikzpicture}
}}
\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& \hooklen{25}{5^2}+\hooklen{16}{4^2}
\tikz[baseline=-0.3ex,very thick]{%
		\useasboundingbox(0,0);%縦0横0の描画領域
		\draw[CUDOrange,dashed](-3.1em,-13ex)
			rectangle ++(3.2em,15ex);
		\draw[CUDMagenta,double,thin](1em,-13ex) 
			rectangle ++(8.6em,15ex);
		}
	-2\times5\times4\times\cos120^\circ \\\pause 
x^2 &=& 25+16 -2\times5\times4\times\hooklen{\cos120^\circ}
	{\Bigl(-\,\myfrac{\,1\,}{2}\Bigr)} \\\pause
x^2 &=& \hooklen{25+16}{41} \textcolor{CUDRed}{{}+{}}{2\hskip-16pt
	\tikz[baseline=1pt] \draw[line width=2.4pt,CUDRed] (0pt,0pt)
		--(16pt,16pt);}\times5
	\times4\times\hooklen{\cos120^\circ}{\Bigl(\quad\myfrac{\,1\,}
		{{2\hskip-16pt\tikz[baseline=1pt]
	\draw[line width=2.4pt,CUDRed] (0pt,0pt)--(16pt,16pt);}}\Bigr)} \\
							\pause
x^2 &=& \hooklen{25+16}{41} + \phantom{2\times{}}\hooklen{5\times4}{20} \\
							\pause
x^2 &=& \hskip8ex	61
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{辺\(x\)の長さを求めなさい}
\vspace*{-2zh}
\begin{eqnarray*}
x^2 &=& 61 \\\pause
x &=& \newpm\sqrt{61}
\end{eqnarray*}

\qquad \(x>0\)なので \pause \hspace*{4ex}\(x=\sqrt{61}\)\quad\□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\end{document}


