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\lfoot{ {\scriptsize{\textsf{\textsc{Stéphane Le Méteil }}}}}


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\begin{document}%\everymath{\displaystyle}
	
\exo Calculer sans utiliser le théorème de Pythagore

\medskip

\begin{tabular}{|c|c|}
\hline
1)\begin{minipage}{0.23\linewidth}\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=5pt 0,linewidth=0.8pt,arrowsize=3pt 2,arrowinset=0.25}
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\begin{minipage}{0.23\linewidth}Calculer $BC$\\ Calculer $AC$\end{minipage}
&
2)\begin{minipage}{0.23\linewidth}
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\begin{minipage}{0.23\linewidth}Calculer $BC$\\ Calculer $AB$\end{minipage} \\ \hline

\hline
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\begin{minipage}{0.23\linewidth}Calculer $AB$\\ Calculer $AC$\end{minipage}
&
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\begin{minipage}{0.23\linewidth}Calculer $\widehat {BAC}$\\ Calculer $AC$\end{minipage} \\ \hline

\hline
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\begin{minipage}{0.23\linewidth}Calculer $\widehat{BAC}$\\ Calculer $BC$\end{minipage}
&
6)\begin{minipage}{0.23\linewidth}
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\begin{minipage}{0.23\linewidth}Calculer $\widehat {BAC}$\\ Calculer $AB$\end{minipage} \\ \hline

\end{tabular}

\vskip 1cm

\exo

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\psdots[dotstyle=*,linecolor=blue](5.,1.)\rput[bl](5.2,0.6){\blue{$C$}}
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\end{minipage}\vrule\qquad
\begin{minipage}{0.53\linewidth}ABCD est un carré de côté 1\\Calculer $AC$ à l'aide du théorème de Pythagore\\Calculer $\widehat {DAC}$\\ En déduire les valeurs exactes de $\cos(45\deg)$ et de $\sin(45\deg)$\end{minipage}

\vskip 1cm

\exo

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\end{minipage}\vrule\qquad
\begin{minipage}{0.53\linewidth}ABC est un triangle équilatéral de côté 1\\Calculer $CH$ à l'aide du théorème de Pythagore\\Calculer $\widehat {CAH}$\\ En déduire les valeurs exactes de $\cos(30\deg)$, de $\sin(30\deg)$, de $\cos(60\deg)$ et de $\sin(60\deg)$\end{minipage}

\end{document}
