\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}
%\usetikzlibrary{intersections,patterns,angles,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\kern-2pt I \quad 2年期末考査勉強会 11月17日}\hfill
{\Large\gtfamily\bfseries 数学I \quad 三角比復習}\hfill
年\hskip8ex 組\hskip8ex 号 \hskip5ex

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

\namae

\vskip-5ex

{\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
$\bullet$ \(\tan\mbox{A}={\,縦\, \over 横} \)

\unitlength=1mm

\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｓｉｎ
\draw(0.5,0)--(2.5,1.2)--(2.5,0)--(0.5,0)--cycle;
\draw(2.3,0.2) rectangle (2.5,0);
\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);
\draw(0.4,-0.2) node{$A$};
\end{tikzpicture}
%
\hspace*{2.4zw}
%
\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｃｏｓ
\draw(0.5,0)--(2.5,1.2)--(2.5,0)--(0.5,0)--cycle;
\draw(2.3,0.2) rectangle (2.5,0);
\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);
\draw(0.4,0.2) node{$A$};
\end{tikzpicture}
%
\hspace*{2.4zw}
%
\begin{tikzpicture}[>=stealth]% 筆記体を書く　ｔａｎ
\draw(0.5,0)--(2.5,1.2)--(2.5,0)--(0.5,0)--cycle;
\draw(2.3,0.2) rectangle (2.5,0);
\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);
\draw(0.4,0.2) node{$A$};
\end{tikzpicture}


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



\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}[rotate=0, scale=0.2,baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize B}] (B) at (12,5);
\coordinate [label=right:{\footnotesize C}]  (C) at (12,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (48pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-1,0)-- ++(0,1) -- ++(1,0);
\draw[dashed] (A) to[out=-30,in=210] node [inner sep=2pt, fill=white]{\small $12$} (C);
\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $5$} (B);
\draw[dashed] (A) to[out=45,in=180] node [inner sep=2pt, fill=white]{\small $13$} (B);
%\draw(6,-1) node[inner sep=1pt]{\small $12$};
%\draw(13,2.5) node[inner sep=1pt]{\small $5$};
\end{tikzpicture}\kotae{{5\over13},{12\over13},{5\over12}}
}{\subtoi 
\begin{tikzpicture}[rotate=-90, scale=0.5, baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=above:{\footnotesize A}] (A) at (0,0);
%\pgfmathparse{sqrt(7)}
\coordinate [label=below:{\footnotesize B}] (B) at ($ ({sqrt(7)},3) $);% (\pgfmathresult,3);
\coordinate [label=below:{\footnotesize C}] (C) at ($ ({sqrt(7)},0) $);% (\pgfmathresult,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (16pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.4,0)-- ++(0,0.4) -- ++(0.4,0);
\draw[dashed] (A) to[out=-60,in=-125] node [inner sep=2pt, fill=white]{\small $\sqrt{7}$} (C);
\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $3$} (B);
\draw[dashed] (A) to[out=80,in=200] node [inner sep=2pt, fill=white]{\small $4$} (B);
%\draw(1.3,-0.7) node[inner sep=1pt]{\small $\sqrt{7}$};
%\draw(3.1,1.5) node[inner sep=1pt]{\small $3$};
\end{tikzpicture}\kotae{{3\over4},{\sqrt{7}\over4},{3\over\sqrt{7}}}
}

\vfill

\def\sincostan{\scriptsize \(\sin A =\raisebox{0mm}{\fbox{\rule[-4mm]{0mm}{10mm}\hskip 6 mm}}\ ,\ 
\cos A =\raisebox{0mm}{\fbox{\rule[-4mm]{0mm}{10mm}\hskip 6 mm}}\ ,\ 
\tan A =\raisebox{0mm}{\fbox{\rule[-4mm]{0mm}{10mm}\hskip 6 mm}}\)}

\nidangumi{\sincostan}{\sincostan}

\nidangumi{\subtoi 
\begin{tikzpicture}[rotate=0, scale=0.5,baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize B}] (B) at (2,5);
\coordinate [label=right:{\footnotesize C}]  (C) at (2,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (12pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.5,0)-- ++(0,0.5) -- ++(0.5,0);
\draw[dashed] (A) to[out=-45,in=225] node [inner sep=2pt, fill=white]{\small $2$} (C);
\draw[dashed] (C) to[out=60,in=-60] node [inner sep=2pt, fill=white]{\small $5$} (B);
\draw[dashed] (A) to[out=90,in=225] node [inner sep=2pt, fill=white]{\small $\sqrt{29}$} (B);
%\draw(1,-0.5) node[inner sep=1pt]{\small $2$};
%\draw(2.5,2.5) node[inner sep=1pt]{\small $5$};
\end{tikzpicture}\kotae{
\ifyurika
	{5\sqrt{29}\over29},{2\sqrt{29}\over29},{5\over2}
\else
	{5\over\sqrt{29}},{2\over\sqrt{29}},{5\over2}
\fi}
}{\subtoi 
\begin{tikzpicture}[rotate=-50, scale=0.3, baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=above:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize B}] (B) at (7,6);
\coordinate [label=below:{\footnotesize C}] (C) at (7,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (36pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.7,0)-- ++(0,0.7) -- ++(0.7,0);
\draw[dashed] (A) to[out=-30,in=210] node [inner sep=2pt, fill=white]{\small $7$} (C);
\draw[dashed] (C) to[out=60,in=-60] node [inner sep=2pt, fill=white]{\small $6$} (B);
\draw[dashed] (A) to[out=80,in=200] node [inner sep=2pt, fill=white]{\small $\sqrt{85}$} (B);
%\draw(3.5,-0.7) node[inner sep=1pt]{\small $7$};
%\draw(7.5,3) node[inner sep=1pt]{\small $6$};
\end{tikzpicture}\kotae{
\ifyurika
	{6\sqrt{85}\over85},{7\sqrt{85}\over85},{6\over7}
\else
	{6\over\sqrt{85}},{7\over\sqrt{85}},{6\over7}
\fi}
}

\vfill


\nidangumi{\sincostan}{\sincostan}


\nidangumi{\subtoi 
\begin{tikzpicture}[rotate=-196, scale=0.16,baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=left:{\footnotesize A}] (A) at (24,7);
\coordinate [label=right:{\footnotesize B}] (B) at (0,0);
\coordinate [label=above:{\footnotesize C}]  (C) at (24,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (48pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-1,0)-- ++(0,1) -- ++(1,0);
\draw[dashed] (B) to[out=-20,in=200] node [inner sep=2pt, fill=white]{\small $24$} (C);
\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $7$} (A);
\draw[dashed] (A) to[out=170,in=40] node [inner sep=2pt, fill=white]{\small $25$} (B);
%\draw(12,-2) node[inner sep=1pt]{\small $24$};
%\draw(25,3.5) node[inner sep=1pt]{\small $7$};
\end{tikzpicture}\\\kotae{{24\over25},{7\over25},{24\over7}}
}{\subtoi 
\begin{tikzpicture}[rotate=-40, scale=0.7, baseline={([yshift=-11pt] current bounding box.north)}]
%\pgfmathparse{sqrt(5)}
\coordinate [label=right:{\footnotesize A}] (A) at ($ ({sqrt(5)},2) $);% (\pgfmathresult,2);
\coordinate [label=left:{\footnotesize B}] (B) at (0,0);
\coordinate [label=below:{\footnotesize C}] (C) at ($ ({sqrt(5)},0) $);% (\pgfmathresult,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (12pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.3,0)-- ++(0,0.3) -- ++(0.3,0);
\draw[dashed] (B) to[out=-60,in=240] node [solid, inner sep=2pt, fill=white]{\small $\sqrt{5}$} (C);
\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $2$} (A);
\draw[dashed] (B) to[out=70,in=190] node [inner sep=2pt, fill=white]{\small $3$} (A);
%\draw(1.2,-0.5) node[inner sep=1pt]{\small $\sqrt{5}$};
%\draw(2.5,1) node[inner sep=1pt]{\small $2$};
\end{tikzpicture}\kotae{{\sqrt{5}\over3},{2\over3},{\sqrt{5}\over2}}
}

\vfill


\nidangumi{\sincostan}{\sincostan}


\ifnum\make_pdf_with_insatsu>0
	\rotatebox{0}{\scriptsize
	{\normalsize\ajKaku{1}}~\ajKakko{1}~\( {5\over13},{12\over13},{5\over12} \)\ 
\ifyurika
	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\sqrt{7}\over7} \)\ 
	\ajKakko{3}~\( {5\sqrt{29}\over29},{2\sqrt{29}\over29},{5\over2} \)\ 
	\ajKakko{4}~\( {6\sqrt{85}\over85},{7\sqrt{85}\over85},{6\over7} \)\ 
	\ajKakko{5}~\( {24\over25},{7\over25},{24\over7} \)\ 
	\ajKakko{6}~\( {\sqrt{5}\over3},{2\over3},{\sqrt{5}\over2} \)\ 
	{\normalsize\ajKaku{2}}~\( \sin30^\circ={1\over2},
	\sin45^\circ={1\over\sqrt{2}},\sin60^\circ={\sqrt{3}\over2},
	{\sqrt{3}\over2},{1\over\sqrt{2}},{1\over2},
	{1\over\sqrt{3}},1,\sqrt{3} \)\ 
	{\normalsize\ajKaku{3}}~\ajKakko{1}~\( \sqrt{21} \)\ 
	\ajKakko{2}~\( {2\over5},{\sqrt{21}\over5},{2\sqrt{21}\over21} \)\ 
	{\normalsize\ajKaku{4}}~\ajKakko{1}~\( \sqrt{7} \)\ 
	\ajKakko{2}~\( {\sqrt{7}\over4},{3\over4},{\sqrt{7}\over3} \)
\else
	\ajKakko{2}~\( {3\over4},{\sqrt{7}\over4},{3\over\sqrt{7}} \)\ 
	\ajKakko{3}~\( {5\over\sqrt{29}},{2\over\sqrt{29}},{5\over2} \)\ 
	\ajKakko{4}~\( {6\over\sqrt{85}},{7\over\sqrt{85}},{6\over7} \)\ 
	\ajKakko{5}~\( {24\over25},{7\over25},{24\over7} \)\ 
	\ajKakko{6}~\( {\sqrt{5}\over3},{2\over3},{\sqrt{5}\over2} \)\ 
	{\normalsize\ajKaku{2}}~\( \sin30^\circ={1\over2},
	\sin45^\circ={1\over\sqrt{2}},\sin60^\circ={\sqrt{3}\over2},
	{\sqrt{3}\over2},{1\over\sqrt{2}},{1\over2},
	{1\over\sqrt{3}},1,\sqrt{3} \)\ 
	{\normalsize\ajKaku{3}}~\ajKakko{1}~\( \sqrt{21} \)\ 
	\ajKakko{2}~\( {2\over5},{\sqrt{21}\over5},{2\over\sqrt{21}} \)\ 
	{\normalsize\ajKaku{4}}~\ajKakko{1}~\( \sqrt{7} \)\ 
	\ajKakko{2}~\( {\sqrt{7}\over4},{3\over4},{\sqrt{7}\over3} \)
\fi
	}
\fi


\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 (A)--(B)--(C)--cycle;
%\draw (1.532,0)--(1.532,0.2)--(1.732,0.2);
\draw (C) ++(-0.2,0)-- ++(0,0.2) -- ++(0.2,0);
%\draw[dashed] (A) to[out=-30,in=-150] node [solid, inner sep=2pt, fill=white]{\small $\sqrt{3}$} (C);
%\draw[dashed] (A) to[out=60,in=180] node [inner sep=2pt, fill=white]{\small $2$} (B);
%\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $1$} (B);
\draw(8pt,0) arc [start angle=0, end angle=30,x radius=8pt, y radius=8pt];
\draw(10pt,0) arc [start angle=0, end angle=30,x radius=10pt, y radius=10pt];
\draw(20pt,5pt) node{\footnotesize $30^\circ$};
\draw(B) +(0,-5pt) arc [start angle=270, end angle=210,x radius=5pt, y radius=5pt];
\draw(B) +(-5pt,-10pt) node{\footnotesize $60^\circ$};
\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 (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] (A) to[out=-45,in=-135] node [inner sep=2pt, fill=white]{\small $1$} (C);
%\draw[dashed] (A) to[out=90,in=180] node [solid, inner sep=1pt, fill=white]{\small $\sqrt{2}$} (B);
%\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $1$} (B);
\draw(5pt,0) arc [start angle=0, end angle=45,x radius=5pt, y radius=5pt];
\draw(12pt,5pt) node{\footnotesize $45^\circ$};
\draw(B) +(0,-5pt) arc [start angle=270, end angle=225,x radius=5pt, y radius=5pt];
\draw(B) +(-5pt,-13pt) node{\footnotesize $45^\circ$};
\end{tikzpicture}
%
\hspace*{7zw}
%
\begin{tikzpicture}[rotate=0, scale=1.2, baseline=(current bounding box.south)]
%\coordinate [label=above:{\footnotesize B}] (B) at (1,1.732);
\coordinate (B) at ($ (1,{sqrt(3)}) $);% (1,1.732);
\coordinate (A) at (0,0);
\coordinate (C) at (1,0);
\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] (A) to[out=-45,in=-135] node [inner sep=2pt, fill=white]{\small $1$} (C);
%\draw[dashed] (A) to[out=90,in=210] node [inner sep=2pt, fill=white]{\small $2$} (B);
%\draw[dashed] (C) to[out=45,in=-45] node [inner sep=2pt, fill=white]{\small $\sqrt{3}$} (B);
\draw(5pt,0) arc [start angle=0, end angle=60,x radius=5pt, y radius=5pt];
\draw(12pt,5pt) node{\footnotesize $60^\circ$};
\draw(B) +(0,-8pt) arc [start angle=270, end angle=240,x radius=8pt, y radius=8pt];
\draw(B) +(0,-10pt) arc [start angle=270, end angle=240,x radius=10pt, y radius=10pt];
\draw(B) +(-5pt,-20pt) node{\footnotesize $30^\circ$};
\end{tikzpicture}

\bigskip
%\vskip7mm

\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{}


\vskip1zh

\kotae{\sin30^\circ={1\over2},\sin45^\circ={1\over\sqrt{2}},
\sin60^\circ={\sqrt{3}\over2},\cos30^\circ={\sqrt{3}\over2},
\cos45^\circ={1\over\sqrt{2}},\cos60^\circ={1\over2}}

\kotae{\tan30^\circ={1\over\sqrt{3}},\tan45^\circ=1,\tan60^\circ=\sqrt{3} }


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

	\subtoi AC の長さを求めなさい。（ \(\mbox{AC}=x\)とすると「三平方の
	定理」より\(x^2+2^2=5^2\)となる。）

%%%%%%%%\vskip-2zh%
\hskip3zw
\begin{tikzpicture}[rotate=0, scale=0.5,baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize B}] (B) at ($ ({sqrt(21)},2) $);%(4.58,2);
\coordinate [label=right:{\footnotesize C}]  (C) at ($ ({sqrt(21)},0) $);%(4.58,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (24pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.5,0)-- ++(0,0.5) -- ++(0.5,0);
\draw[dashed] (B) to[out=-45,in=45] node [inner sep=2pt, fill=white]{\small $2$} (C);
\draw[dashed] (A) to[out=60,in=180] node [inner sep=2pt, fill=white]{\small $5$} (B);
%\draw(45:2.5) node[inner sep=1pt]{\small $5$};
%\draw(5,1) node[inner sep=1pt]{\small $2$};
\end{tikzpicture}\kotae{\sqrt{21}}

\vfill

	\subtoi \(\sin A, \cos A, \tan A\)の値を求めなさい。
\hfill\sincostan

\kotae{{2\over5},{\sqrt{21}\over5},
\ifyurika
	{2\sqrt{21}\over21}
\else
	{2\over\sqrt{21}}
\fi}

 \vskip3zh

\normalsize
\toi 次の直角三角形ABCで，\(\sin A, \cos A, \tan A\)の値を求めたい。

	\subtoi BC の長さを求めなさい。%（ \(\mbox{BC}=x\)とすると「三平方の
					%	定理」より\(3^2+x^2=4^2\)となる。）

%%%%%%%%\vskip-2zh%
\hskip3zw
\begin{tikzpicture}[rotate=0, scale=0.5,baseline={([yshift=-11pt] current bounding box.north)}]
\coordinate [label=left:{\footnotesize A}] (A) at (0,0);
\coordinate [label=right:{\footnotesize B}] (B) at ($ (3,{sqrt(7)}) $);%(3,2.64);
\coordinate [label=right:{\footnotesize C}]  (C) at (3,0);
\begin{scope}
\clip (A)--(B)--(C)--cycle;
\draw (A) circle (16pt);
\end{scope}
\draw (A)--(B)--(C)--cycle;
\draw (C) ++(-0.5,0)-- ++(0,0.5) -- ++(0.5,0);
\draw[dashed] (A) to[out=80,in=190] node [inner sep=2pt, fill=white]{\small $4$} (B);
\draw[dashed] (A) to[out=-45,in=225] node [solid, inner sep=2pt, fill=white]{\small $3$} (C);
%\draw(55:2) node[inner sep=1pt]{\small $4$};
%\draw(1.5,-0.5) node[inner sep=1pt]{\small $3$};
\end{tikzpicture}\kotae{\sqrt{7}}

\vfill

\subtoi \(\sin A, \cos A, \tan A\)の値を求めなさい。 
\hfill\sincostan

\kotae{{\sqrt{7}\over4},{3\over4},{\sqrt{7}\over3}}

\ifnum\make_pdf_with_insatsu>0
	\vskip3zh\mbox{}
\fi

\end{document}
