2015-2016/3e/Geometrie/Trigonometrie/intro_trigo.tex
2017-06-16 09:48:54 +03:00

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\documentclass[a4paper,12pt]{/media/documents/Cours/Prof/Enseignements/tools/style/classExo}
\usepackage{/media/documents/Cours/Prof/Enseignements/2015_2016}
% Title Page
\titre{Introduction à la trigonométrie}
\classe{Troisième}
\date{mars 2014}
\pagestyle{empty}
\geometry{left=5mm,right=5mm, bottom= 10mm, top=10mm}
\begin{document}
Compléter le tableau suivant en fonction des triangles.
\begin{tabular}{|c|*{6}{c|}}
\hline
Triangle & Hypoténuse & coté adjacent & coté opposé & angle & $\frac{\mbox{coté adjacent}}{\mbox{Hypoténuse}}$ & $\frac{\mbox{coté opposé}}{\mbox{Hypoténuse}}$ \\
\hline
ABC & ...... = ...... & ...... = ...... & ...... = ...... & ...... = ...... & $\frac{......}{......} = ......$ & $\frac{......}{......} = ......$ \\
\hline
DEF & ...... = ...... & ...... = ...... & ...... = ...... & ...... = ...... & $\frac{......}{......} = ......$ & $\frac{......}{......} = ......$ \\
\hline
GHI & ...... = ...... & ...... = ...... & ...... = ...... & ...... = ...... & $\frac{......}{......} = ......$ & $\frac{......}{......} = ......$ \\
\hline
JKL & ...... = ...... & ...... = ...... & ...... = ...... & ...... = ...... & $\frac{......}{......} = ......$ & $\frac{......}{......} = ......$ \\
\hline
MNP & ...... = ...... & ...... = ...... & ...... = ...... & ...... = ...... & $\frac{......}{......} = ......$ & $\frac{......}{......} = ......$ \\
\hline
\end{tabular}
\begin{center}
\includegraphics[scale=0.4]{./fig/triangles}
\end{center}
\begin{Exo}
\textbf{Avec la figure suivante, calculer la longueur $BA$.}
\\[0.3cm]
\begin{minipage}[h]{0.2\textwidth}
\includegraphics[scale=0.15]{./fig/triangleABC}
\end{minipage}
\begin{minipage}[h]{0.8\textwidth}
\textit{Questions à se poser}
\begin{itemize}
\item On connait
\begin{center}
hypoténuse\hspace{0.5cm} opposé \hspace{0.5cm}adjacent \hspace{0.5cm}angle
\end{center}
\item On cherche
\begin{center}
hypoténuse\hspace{0.5cm} opposé \hspace{0.5cm}adjacent \hspace{0.5cm}angle
\end{center}
\end{itemize}
\begin{itemize}
\item On utilise la formule
\begin{eqnarray*}
\cos( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.5cm}
\sin( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.5cm}
\tan( \widehat{BAC}) = \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}}
\end{eqnarray*}
\end{itemize}
\end{minipage}
\textit{Rédaction:}
\noindent\fbox{\parbox{\linewidth-2\fboxrule-2\fboxsep}{
.\\
\parbox{1cm}{\dotfill} est un triangle rectangle en \parbox{0.5cm}{\dotfill} donc
\begin{eqnarray*}
\parbox{1cm}{\dotfill}(\widehat{BAC}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les lettres)}\\[0.3cm]
\parbox{1cm}{\dotfill}(\parbox{1cm}{\dotfill}) &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
\parbox{1cm}{\dotfill} &=& \frac{\parbox{2cm}{\dotfill}}{\parbox{2cm}{\dotfill}} \hspace{0.7cm} \textit{(avec les chiffres)} \\[0.3cm]
BA & = & \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} = \parbox{1cm}{\dotfill}
\end{eqnarray*}
Donc $BA = \parbox{1cm}{\dotfill}$
}}
\end{Exo}
\end{document}
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