%
%	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,angles}
%\usetikzlibrary{arrows.meta,patterns,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=0pt
%■ \( \log\log \)でlogとlogの間に挿入される細スペース（省略形は \,）
%\thinmuskip=3mu%■default
%■ \( a+b \)で+の前後に挿入される中スペース（省略形は \>）
%\medmuskip=4mu plus 2mu minus 4mu%■default
\medmuskip=2mu plus 1mu 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%■default
\thickmuskip=3mu plus 2mu minus 2mu
% mu は math units の省略で長さの単位

\newcommand{\henao}{\tikz[baseline=-2pt]\draw(3pt,3pt)
		node[circle, fill=CUDBlue,
			inner sep=1pt]{\small\textcolor{white}{\bf 辺}};
}

%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■



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

\begin{frame}[t]
\frametitle{\ajKakko{1}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-3ex}
\begin{tikzpicture}[rotate=0, scale=0.8, line join=round,very thick,
	>=stealth, baseline={(current bounding box.north)}]
%\draw[help lines] (-2,-1) grid (16,7.2);
\useasboundingbox (-2,-1) rectangle (16,7.2);
\alt<3->{\coordinate [label=left:\textcolor{CUDRed}{$A$}] (A) at (0,0);}
	{\coordinate [label=left:$A$] (A) at (0,0);}
\coordinate (B) at (8,0);
\coordinate (C) at (60:5);

\visible<2->{\draw(7.5,6) node{\small
\( \raisebox{-9pt}{\textcolor{CUDGreen}{\shortstack{向かい\\[-3pt]の辺}}}^2 =
	\colorbox{CUDPink}{\( \henao^2 \!+\henao^2 \)}-2\times\!
	\colorbox{CUDCream}{\( \henao\!\times\!\henao \)}
	\!\times\!\cos \text{\textcolor{CUDRed}{間の角度}} \) };
	}
\visible<3->{\pic[fill=CUDRed, angle radius=20pt]{angle=B--A--C};}

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=20pt]{angle=B--A--C};

\alt<3->{\draw[dashed] (A) to[out=90,in=200]
		node[CUDBlue,inner sep=6pt, fill=white]{$5$} (C);
	\draw[dashed] (A) to[out=-20,in=200]
		node [CUDBlue,inner sep=6pt, fill=white]{$8$} (B);
	\draw[dashed] (C) to[out=0,in=110]
		node [CUDGreen,inner sep=6pt, fill=white]{$7$} (B);
	\draw[CUDBlue,line width=3pt](B)--(A)--(C);
	}
	{\draw[dashed] (A) to[out=90,in=200]
		node[inner sep=6pt, fill=white]{$5$} (C);
	\draw[dashed] (A) to[out=-20,in=200]
		node [inner sep=6pt, fill=white]{$8$} (B);
	\draw[dashed] (C) to[out=0,in=110]
		node [inner sep=6pt, fill=white]{$7$} (B);
	}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-3ex}
\begin{tikzpicture}[rotate=0, scale=0.8, line join=round, very thick, >=stealth,
	baseline={(current bounding box.north)}]
%\draw[help lines] (-2,-1) grid (16,7.2);
\useasboundingbox (-2,-1) rectangle (16,7.2);
\coordinate [label=left:{\textcolor{CUDRed}{$A$}}] (A) at (0,0);
\coordinate (B) at (8,0);
\coordinate (C) at (60:5);

\alt<2->{\draw(7.5,6) node{\( \textcolor{CUDGreen}{7}^2 =
	\colorbox{CUDPink}{%
		\( \textcolor{CUDBlue}{8}^2+\textcolor{CUDBlue}{5}^2 \)}
	-2\times\colorbox{CUDCream}{%
		\( \textcolor{CUDBlue}{8}\times\textcolor{CUDBlue}{5} \)}
		\times\cos \textcolor{CUDRed}{A} \)};
	}
	{\draw(7.5,6) node{\( \textcolor{CUDGreen}{7}^2 =
	\colorbox{CUDPink}{%
		\( \textcolor{CUDBlue}{5}^2+\textcolor{CUDBlue}{8}^2 \)}
	-2\times\colorbox{CUDCream}{%
	\( \textcolor{CUDBlue}{5}\times\textcolor{CUDBlue}{8} \)}
	\times\cos \textcolor{CUDRed}{A} \)};
	}
\pic[fill=CUDRed, angle radius=20pt]{angle=B--A--C};

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=20pt]{angle=B--A--C};

\draw[dashed] (A) to[out=90,in=200]
	node[CUDBlue,inner sep=6pt, fill=white]{$5$} (C);
\draw[dashed] (A) to[out=-20,in=200]
	node [CUDBlue,inner sep=6pt, fill=white]{$8$} (B);
\draw[dashed] (C) to[out=0,in=110]
	node [CUDGreen,inner sep=6pt, fill=white]{$7$} (B);
\draw[CUDBlue,line width=3pt](B)--(A)--(C);

\visible<2->{\draw(12,3.5) node{でもOKです。};
	\draw(12,2.3) node[CUDBlue]{$5$と$8$がペアが重要};
	}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-45pt}
\begin{eqnarray*}
7^2 &=& \hooklen{25}{5^2}+\hooklen{64}{8^2}
	-2\times5\times8\times\cos A \\ \pause
49 &=& 25+64-\hooklen{2\times5\times8}{80}\times
	\alt<3->{\textcolor{CUDRed}{\cos A}}{\cos A} \\ \pause
49 &=& \hooklen{25+64}{89}
	-\hooklen{2\times5\times8\times\cos A}{80\,\textcolor{CUDRed}{x}} \\
   & & \kern10ex \textcolor{CUDRed}{(\cos A=x\text{とおく})}\\ \pause
80\,\textcolor{CUDRed}{x} &=& 89-49 \\ \pause
80\,x &=& \hooklen{89-49}{40}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-5ex}
\begin{eqnarray*}
80\,x &=& 40 \\[1ex] \pause
\myfrac{80\,x }{\textcolor{CUDRed}{80}} &=&
	\myfrac{\,40\,}{\textcolor{CUDRed}{80}} \\[1ex] \pause
\hooklen{80\,x}{x} &=& \myfrac{\,1\,}{2} \\[1ex] \pause
\cos A &=& \myfrac{\,1\,}{2} \qquad (\cos A=x\ \text{を元に戻す})
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\quad \(A\)の角度を求めなさい（余弦定理）}

\( \cos A = \myfrac{\,1\,}{2} \)となる\(A\)を考える。
\visible<3->{\□答\(A=60^\circ\)}\\[1ex]

\fbox{\begin{minipage}[t]{0.96\textwidth}
{\footnotesize これらの値を頭の中で思い浮かべる\\[1ex]
\centerline{\( \cos 30^\circ=\myfrac{\sqrt{3}}{2}\hskip7ex
	\cos45^\circ=\myfrac{1}{\sqrt{2}}\hskip7ex
	\alt<2->{\,\textcolor{CUDRed}{\cos60^\circ=\myfrac{\,1\,}{2}}}
		{\cos60^\circ=\myfrac{\,1\,}{2}}\)}
\centerline{\( \cos 120^\circ=\myfrac{\,-1\,}{2}\hskip4ex
	\cos135^\circ=\myfrac{-1}{\sqrt{2}}\hskip4ex
	\cos150^\circ=\myfrac{-\sqrt{3}}{2}\)}
	}
\end{minipage}}
\end{frame}
%%%%%%%%%%%%%%%%%%%　■■■■　%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\quad \(C\)の角度を求めなさい（余弦定理）}
\vspace*{-2ex}
\begin{tikzpicture}[rotate=0, scale=1.2, line join=round, very thick,
	>=stealth, baseline={([yshift=-16pt] current bounding box.north)}]
%\draw[help lines] (-2,-1) grid (10,4.5);
\useasboundingbox (-2,-1) rectangle (10,4.5);
\coordinate (B) at (0,0);
\coordinate (A) at ($ ({5+3*cos(60)},{3*sin(60)}) $);
\alt<3->{\coordinate [label=below right:\textcolor{CUDRed}{$C$}] (C) at (5,0);}
	{\coordinate [label=below right:$C$] (C) at (5,0);}

\visible<2->{\draw(4,3.8) node{\small
\( \raisebox{-9pt}{\textcolor{CUDGreen}{\shortstack{向かい\\[-3pt]の辺}}}^2 =
	\colorbox{CUDPink}{\( \henao^2 \!+\henao^2 \)}-2\times\!
	\colorbox{CUDCream}{\( \henao\!\times\!\henao \)}
	\!\times\!\cos \text{\textcolor{CUDRed}{間の角度}} \) };
	}
\visible<3->{\pic[fill=CUDRed, angle radius=16pt]{angle=A--C--B};}

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=16pt]{angle=A--C--B};

\alt<3->{\draw[dashed] (A) to[out=-80,in=20]
		node [CUDBlue,inner sep=3pt, fill=white]{$3$} (C);
	\draw[dashed] (A) to[out=180,in=45]
		node [CUDGreen, inner sep=6pt, fill=white]{$7$} (B);
	\draw[dashed] (C) to[out=210,in=-30]
		node [CUDBlue, inner sep=6pt, fill=white]{$5$} (B);
	\draw[CUDBlue,line width=3pt](B)--(C)--(A);
	}{\draw[dashed] (A) to[out=-80,in=20]
		node [inner sep=3pt, fill=white]{$3$} (C);
	\draw[dashed] (A) to[out=180,in=45]
		node [inner sep=6pt, fill=white]{$7$} (B);
	\draw[dashed] (C) to[out=210,in=-30]
		node [inner sep=6pt, fill=white]{$5$} (B);
	}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\quad \(C\)の角度を求めなさい（余弦定理）}
\vspace*{-2ex}
\begin{tikzpicture}[rotate=0, scale=1.2, line join=round, very thick,
	>=stealth, baseline={([yshift=-16pt] current bounding box.north)}]
%\draw[help lines] (-2,-1) grid (10,4.5);
\useasboundingbox (-2,-1) rectangle (10,4.5);
\coordinate [label=below right:\textcolor{CUDRed}{$C$}] (C) at (5,0);
\coordinate (B) at (0,0);
\coordinate (A) at ($ ({5+3*cos(60)},{3*sin(60)}) $);
\draw(4,3.8) node{\( \textcolor{CUDGreen}{7}^2 =
	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{5}^2+
	\textcolor{CUDBlue}{3}^2 \)}
	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{5}\times
	\textcolor{CUDBlue}{3} \)}\times\cos \textcolor{CUDRed}{C} \) };

\pic[fill=CUDRed, angle radius=16pt]{angle=A--C--B};

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=16pt]{angle=A--C--B};

\draw[dashed] (A) to[out=-80,in=20]
	node [CUDBlue,inner sep=3pt, fill=white]{$3$} (C);
\draw[dashed] (A) to[out=180,in=45]
	node [CUDGreen, inner sep=6pt, fill=white]{$7$} (B);
\draw[dashed] (C) to[out=210,in=-30]
	node [CUDBlue, inner sep=6pt, fill=white]{$5$} (B);
\draw[CUDBlue,line width=3pt](B)--(C)--(A);
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\quad \(C\)の角度を求めなさい（余弦定理）}
\vspace*{-5ex}
\begin{eqnarray*}
7^2 &=&	\hooklen{25}{5^2}+3^2
	-2\times5\times3\times\cos C \\[6pt] \pause
49 &=& 25+\hooklen{3^2}{9}-\hooklen{2\times5\times3}{30}\times
	\alt<3->{\textcolor{CUDRed}{\cos C}}{\cos C} \\ \pause
49 &=& \hooklen{25+3^2}{34}
	-\hooklen{2\times5\times3\times\cos C}{30\,\textcolor{CUDRed}{x}} \\
   & & \kern10ex \textcolor{CUDRed}{(\cos C=x\text{とおく})}\\ \pause
30\,\textcolor{CUDRed}{x} &=& 34-49 \\ \pause
30\,x &=& \hooklen{34-49}{-15}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\quad \(C\)の角度を求めなさい（余弦定理）}
\vspace*{-5ex}
\begin{eqnarray*}
30\, x &=& -15 \\[6pt] \pause
\myfrac{30\, x}{\textcolor{CUDRed}{30}}&=&
	\myfrac{-15}{\textcolor{CUDRed}{30}} \\[6pt] \pause
\hooklen{30\, x}{x} &=& \myfrac{\, -1\,}{2} \\[6pt] \pause
\cos C &=& \myfrac{\, -1\,}{2} \qquad (\cos C=x\ \text{を元に戻す})
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\quad \(C\)の角度を求めなさい（余弦定理）}
\( \cos C = \myfrac{-1}{2} \)となる\(C\)を考える。
\visible<3->{\□答\(C=120^\circ\)}\\[1ex]

\fbox{\begin{minipage}[t]{0.96\textwidth}
{\footnotesize これらの値を頭の中で思い浮かべる\\[1ex]
\centerline{\( \cos 30^\circ=\myfrac{\sqrt{3}}{2}\hskip7ex 
\cos45^\circ=\myfrac{1}{\sqrt{2}}%
	\hskip7ex \cos60^\circ=\myfrac{\,1\,}{2}\)}
\centerline{\( \alt<2->{\textcolor{CUDRed}{\cos120^\circ=\myfrac{\,-1\,}{\,2\,}}}
		{\cos120^\circ=\myfrac{\,-1\,}{\,2\,}}\hskip3ex
	\cos135^\circ=\myfrac{-1}{\sqrt{2}}%
	\hskip4ex \cos150^\circ=\myfrac{-\sqrt{3}}{2}\)}
	}
\end{minipage}}
\end{frame}
%%%%%%%%%%%%%%%%%%%　■■■■　%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-3ex}
\begin{tikzpicture}[rotate=0, scale=2.8, line join=round, very thick,
	>=stealth, baseline={(current bounding box.north)}]
%\draw[help lines] (0,-0.5) grid (5,2);
\useasboundingbox (0,-0.5) rectangle (5,2);
\coordinate (B) at (0,0);
\coordinate (C) at ($ ({sqrt(5)},0) $);
\alt<3->{\coordinate [label=right:\textcolor{CUDRed}{$A$}](A)
		at ($ ({sqrt(5)+sqrt(2)*cos(60)},{sqrt(2)*sin(60)}) $);}
	{\coordinate [label=right:$A$](A)
		at ($ ({sqrt(5)+sqrt(2)*cos(60)},{sqrt(2)*sin(60)}) $);}

\visible<2->{\draw(2.5,1.7) node{\small
\( \raisebox{-9pt}{\textcolor{CUDGreen}{\shortstack{向かい\\[-3pt]の辺}}}^2 =
	\colorbox{CUDPink}{\( \henao^2 \!+\henao^2 \)}-2\times\!
	\colorbox{CUDCream}{\( \henao\!\times\!\henao \)}
	\!\times\!\cos \text{\textcolor{CUDRed}{間の角度}} \) };
	}
\visible<3->{\pic[fill=CUDRed, angle radius=24pt]{angle=B--A--C};}

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=24pt]{angle=B--A--C};

\alt<3->{\draw[dashed] (A) to[out=-80,in=10]
		node [CUDBlue,inner sep=2pt, fill=white]{$\sqrt{2}$} (C);
	\draw[dashed] (A) to[out=180,in=45]
		node [CUDBlue, inner sep=6pt, fill=white]{$3$} (B);
	\draw[dashed] (C) to[out=200,in=-20]
		node [CUDGreen, inner sep=6pt, fill=white]{$\sqrt{5}$} (B);
	\draw[CUDBlue,line width=3pt](B)--(A)--(C);
	}{\draw[dashed] (A) to[out=-80,in=10]
		node [inner sep=2pt, fill=white]{$\sqrt{2}$} (C);
	\draw[dashed] (A) to[out=180,in=45]
		node [inner sep=6pt, fill=white]{$3$} (B);
	\draw[dashed] (C) to[out=200,in=-20]
		node [inner sep=6pt, fill=white]{$\sqrt{5}$} (B);
	}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-3ex}
\begin{tikzpicture}[rotate=0, scale=2.8, line join=round, very thick,
	>=stealth, baseline={(current bounding box.north)}]
%\draw[help lines] (0,-0.5) grid (5,2);
\useasboundingbox (0,-0.5) rectangle (5,2);
\coordinate (B) at (0,0);
\coordinate (C) at ($ ({sqrt(5)},0) $);
\coordinate [label=right:\textcolor{CUDRed}{$A$}](A)
		at ($ ({sqrt(5)+sqrt(2)*cos(60)},{sqrt(2)*sin(60)}) $);

\alt<2->{\draw(2.5,1.7) node{
	\( \textcolor{CUDGreen}{\sqrt{5}}^{\kern4pt 2} =
	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{\sqrt{2}}^{\kern4pt 2}+
	\textcolor{CUDBlue}{3}^2 \)}%
	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{\sqrt{2}}\times
	\textcolor{CUDBlue}{3} \)}\times\cos \textcolor{CUDRed}{A} \)};
	}
	{\draw(2.5,1.7) node{
	\( \textcolor{CUDGreen}{\sqrt{5}}^{\kern4pt 2} =
	\colorbox{CUDPink}{\( \textcolor{CUDBlue}{3}^2+
	\textcolor{CUDBlue}{\sqrt{2}}^{\kern4pt 2} \)}%
	-2\times\colorbox{CUDCream}{\( \textcolor{CUDBlue}{3}\times
	\textcolor{CUDBlue}{\sqrt{2}} \)}\times\cos \textcolor{CUDRed}{A}\)};
	}
\pic[fill=CUDRed, angle radius=24pt]{angle=B--A--C};

\draw (A)--(B)--(C)--cycle;
\pic[draw=black, angle radius=24pt]{angle=B--A--C};

\draw[dashed] (A) to[out=-80,in=10]
		node [CUDBlue,inner sep=2pt, fill=white]{$\sqrt{2}$} (C);
	\draw[dashed] (A) to[out=180,in=45]
		node [CUDBlue, inner sep=6pt, fill=white]{$3$} (B);
	\draw[dashed] (C) to[out=200,in=-20]
		node [CUDGreen, inner sep=6pt, fill=white]{$\sqrt{5}$} (B);
\draw[CUDBlue,line width=3pt](B)--(A)--(C);
\visible<2->{\draw(4.3,1.2) node{\small でもOKです。};
	\draw(4.3,0.9) node[CUDBlue]{\small $\sqrt{2}$と$3$がペアが重要};
	}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-42pt}
\begin{eqnarray*}
\sqrt{5}^{\kern4pt 2} &=&
	3^2+\sqrt{2}^{\kern4pt 2}
	-2\times3\times\sqrt{2}\times\cos A \\[6pt] \pause
\hooklen{\sqrt{5}^{\kern4pt 2}}{5} &=&
	\hooklen{3^2}{9}+\hooklen{\sqrt{2}^{\kern4pt 2}}{2}
	-\hooklen{2\times3\times\sqrt{2}}{6\sqrt{2}}\times
	\alt<3->{\textcolor{CUDRed}{\cos A}}{\cos A} \\[6pt] \pause
\hooklen{\sqrt{5}^{\kern4pt 2}}{5} &=&
	\hooklen{3^2+\sqrt{2}^{\kern4pt 2}}{11}
	-\hooklen{2\times3\times\sqrt{2}\times\cos A}{6\sqrt{2}\ 
	\textcolor{CUDRed}{x}}\\[6pt]
  & & \kern10ex \textcolor{CUDRed}{(\cos A=x\text{とおく})}\\[-4pt] \pause
6\sqrt{2}\ \textcolor{CUDRed}{x} &=& 11-5 \\[6pt] \pause
6\sqrt{2}\ x &=& \hooklen{11-5}{6}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\quad \(A\)の角度を求めなさい（余弦定理）}
\vspace*{-5ex}
\begin{eqnarray*}
6\sqrt{2}\ x &=& 6 \\[6pt] \pause
\myfrac{6\sqrt{2}\ x }{\textcolor{CUDRed}{6\sqrt{2}}}&=&
	\myfrac{6}{\textcolor{CUDRed}{6\sqrt{2}}} \\[6pt] \pause
\hooklen{6\sqrt{2}\ x}{x} &=& \myfrac{1}{\sqrt{2}} \\[6pt] \pause
\cos A &=& \myfrac{1}{\sqrt{2}} \qquad (\cos A=x\ \text{を元に戻す})
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\quad \(A\)の角度を求めなさい（余弦定理）}
\( \cos A = \myfrac{1}{\sqrt{2}} \)となる\(A\)を考える。
\visible<3->{\□答\(A=45^\circ\)}\\[1ex]

\fbox{\begin{minipage}[t]{0.96\textwidth}
{\footnotesize これらの値を頭の中で思い浮かべる\\[1ex]
\centerline{\( \cos 30^\circ=\myfrac{\sqrt{3}}{2}\hskip7ex
	\alt<2->{\,\textcolor{CUDRed}{\cos45^\circ=\myfrac{1}{\sqrt{2}}}}
		{\cos45^\circ=\myfrac{1}{\sqrt{2}}}\hskip7ex
		\cos60^\circ=\myfrac{\,1\,}{2}\)}
\centerline{\( \cos 120^\circ=\myfrac{\,-1\,}{2}\hskip4ex
	\cos135^\circ=\myfrac{-1}{\sqrt{2}}%
	\hskip4ex \cos150^\circ=\myfrac{-\sqrt{3}}{2}\)}
	}
\end{minipage}}
\end{frame}
%%%%%%%%%%%%%%%%%%%　■■■■　%%%%%%%%%%%%%%%%%%%%%%%%%%




\begin{frame}[t]
\frametitle{余弦定理}
\vspace*{-4ex}
\[ \cos \text{\textcolor{CUDRed}{間の角度}}=
	\myfrac{\ \colorbox{CUDPink}{\( \henao^2 \!+\henao^2 \)}
	- \text{\textcolor{CUDGreen}{向かいの辺}}^2 \ 
}{2\times\!	\colorbox{CUDCream}{\( \henao\!\times\!\henao \)}}
\]

\vskip2ex
この形に変形して使うこともあります。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}


