121 lines
4.5 KiB
TeX
121 lines
4.5 KiB
TeX
\documentclass[a4paper,10pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
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% Title Page
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\title{Trigonométrie - Exercices}
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\author{}
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\date{}
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\fancyhead[L]{Troisième}
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\fancyhead[C]{\Thetitle}
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\fancyhead[R]{\thepage}
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\begin{document}
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\thispagestyle{empty}
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\begin{center}
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\large{Calculer une longueur}
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\end{center}
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\normalsize
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\begin{Exo}
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\begin{enumerate}
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\item \textbf{Avec la figure suivante, calculer la longueur $BA$.}
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\\[0.3cm]
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\begin{minipage}[h]{0.15\textwidth}
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\includegraphics[scale=0.15]{./fig/triangleABC}
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\end{minipage}
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\begin{minipage}[h]{0.3\textwidth}
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\textit{Questions à se poser}
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\begin{itemize}
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\item On connait
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\begin{center}
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hypoténuse\hspace{0.5cm} opposé \hspace{0.5cm}adjacent \hspace{0.5cm}angle
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\end{center}
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\item On cherche
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\begin{center}
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hypoténuse\hspace{0.5cm} opposé \hspace{0.5cm}adjacent \hspace{0.5cm}angle
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\end{center}
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\end{itemize}
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\end{minipage}
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\begin{itemize}
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\item On utilise la formule
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\begin{eqnarray*}
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\cos( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.5cm}
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\sin( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.5cm}
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\tan( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}}
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\end{eqnarray*}
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\end{itemize}
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\textit{Rédaction:}
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\noindent\fbox{\parbox{\linewidth-2\fboxrule-2\fboxsep}{
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.\\[0.2cm]
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\parbox{1cm}{\dotfill} est un triangle rectangle en \parbox{0.5cm}{\dotfill} donc
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\begin{eqnarray*}
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\parbox{1cm}{\dotfill}(\widehat{BAC}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les lettres)}\\[0.3cm]
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\parbox{1cm}{\dotfill}(\parbox{1cm}{\dotfill}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
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\parbox{1cm}{\dotfill} &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
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BA & = & \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} = \parbox{1cm}{\dotfill}
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\end{eqnarray*}
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Donc $BA = \parbox{1cm}{\dotfill}$
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}}
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\item \textbf{Calculer la longueur $EF$}
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\begin{minipage}[h]{0.15\textwidth}
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\includegraphics[scale=0.2]{./fig/triangleDEF}
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\end{minipage}
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\begin{minipage}[h]{0.35\textwidth}
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\textit{Rédaction:}
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\noindent\fbox{\parbox{\linewidth-2\fboxrule-2\fboxsep-0.1\textwidth}{
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.\\[0.2cm]
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\parbox{1cm}{\dotfill} est un triangle rectangle en \parbox{0.5cm}{\dotfill} donc
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\begin{eqnarray*}
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\parbox{1cm}{\dotfill}(\widehat{EFD}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les lettres)}\\[0.3cm]
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\parbox{1cm}{\dotfill}(\parbox{1cm}{\dotfill}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
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\parbox{1cm}{\dotfill} &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
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EF & = & \frac{ \parbox{1cm}{\dotfill} }{ \parbox{1cm}{\dotfill}} = \parbox{1cm}{\dotfill}
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\end{eqnarray*}
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Donc $EF = \parbox{1cm}{\dotfill}$
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}}
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\end{minipage}
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\end{enumerate}
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\end{Exo}
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\vfill\eject
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\begin{Exo}
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En reprenant la rédaction présenté au dessus, faire les exercices suivants.
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\begin{enumerate}
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\item Calculer la longueur $HI$.
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\begin{center}
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\includegraphics[scale=0.2]{./fig/triangleGHI}
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\end{center}
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\item Calculer la longueur $LK$. On donne $\widehat{JKL} = 10$°.
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\begin{center}
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\includegraphics[scale=0.2]{./fig/triangleJKL}
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\end{center}
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\item $MNO$ est un triangle rectangle en $M$ tel que $OM=3cm$ et $\widehat{OMN} = 60$°.
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\begin{enumerate}
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\item Faire une figure à main levée.
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\item Calculer $NO$.
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\end{enumerate}
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\item $PQR$ est un triangle rectangle en $P$ tel que $RQ=10cm $ et $\widehat{PRQ} = 98$°. Calculer la longueur $QP$.
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\end{enumerate}
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\end{Exo}
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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