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\newcommand{\Ch}{Trigonom\'etrie}
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\debutDmCha



\begin{ExoF}
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Graphiquement, et après avoir éventuellement déplacer certains vecteurs, on lit : 
$$( \overrightarrow{PN}, \overrightarrow{PM}) =- \dfrac{\pi}3 \quad [2\pi]
\qquad \qquad ( \overrightarrow{MN}, \overrightarrow{PN})=-\dfrac{\pi}3 \quad[2\pi]
\qquad \qquad ( \overrightarrow{NM}, \overrightarrow{MP})=-\dfrac{2\pi}3 \quad [2\pi]
 \qquad \qquad ( \overrightarrow{PN}, \overrightarrow{PI})=-\dfrac{\pi}{6} \quad [2\pi]$$
\end{ExoF}




\begin{ExoF}\\
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\begin{enumerate}
 \item $( \overrightarrow{AB}, \overrightarrow{AC})=\dfrac{\pi}{4}$ en mesure principale.\\
 Mais aussi 
 $( \overrightarrow{AB}, \overrightarrow{AC})=\dfrac{\pi}{4}+2\pi=\dfrac{9}{4}$ par exemple.
 \item  On peut, là encore, déplacer les vecteurs, mais utilisons également les propriétés des angles associés pour nous entraîner.
\begin{eqnarray*}
 ( \overrightarrow{CD}, \overrightarrow{CA})&=&\dfrac{\pi}{4}  \quad [2\pi]\\\\
( \overrightarrow{BC}, \overrightarrow{OB})&=&( \overrightarrow{BC}, \overrightarrow{BO})+\pi = \dfrac{\pi}4+\pi = \dfrac{5\pi}{4}=-\dfrac{3\pi}{4} \quad[2\pi]\\\\
( \overrightarrow{DO}, \overrightarrow{AC})&=&( \overrightarrow{DO}, \overrightarrow{OC})=( \overrightarrow{OD}, \overrightarrow{OC})+\pi = -\dfrac{\pi}2 + \pi = \dfrac{\pi}2 \quad [2\pi]\\\\
 ( \overrightarrow{OA}, \overrightarrow{AC})&=&( \overrightarrow{OA}, \overrightarrow{OC})=\pi \quad [2\pi]\\\\
( \overrightarrow{BA}, \overrightarrow{CD})&=&( \overrightarrow{BA}, \overrightarrow{BA})=0\quad [2\pi]\\\\
 ( \overrightarrow{OB}, \overrightarrow{AD})&=&( \overrightarrow{OB}, \overrightarrow{BC})=( \overrightarrow{BO}, \overrightarrow{BC})+\pi = -\dfrac{\pi}4+\pi = \dfrac{3\pi}4 \quad [2\pi]\\\\
( \overrightarrow{OC}, \overrightarrow{OB})&=&-\dfrac{\pi}2\quad [2\pi] \\\\
( \overrightarrow{OC}, \overrightarrow{OA})&=&\pi \quad [2\pi]\\\\
 ( \overrightarrow{DA}, \overrightarrow{CO})&=&( \overrightarrow{CB}, \overrightarrow{CO})=-\dfrac{\pi}{4} \quad [2\pi]  
\end{eqnarray*}
\end{enumerate}
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\end{ExoF}


\newpage

\begin{ExoF}
\begin{enumerate}
 \item \begin{center}
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Remarquons que coder la figure aide largement à répondre aux questions qui suivent ...
\item \begin{enumerate}
\item %D\'emontrer que le triangle ADE est isoc\`ele
On sait que $AEB$ est un triangle équilatéral, donc $AB=AE$.\\
On sait aussi que $ABCD$ est un carré, donc $AB=AD$.\\
On en déduit de suite que $AE=AD$, donc $AED$ est équilatéral en $A$.
\item %D\'emontrer que $ (\overrightarrow{ED}, \overrightarrow{EA})=\dfrac{5\pi}{12}$ 
Dans le triangle $AED$, on sait que les angles de la base sont égaux. En notant ces angles en angles orientés de vecteurs (avec 
la même orientation) on a donc $(\overrightarrow{ED}, \overrightarrow{EA})=(\overrightarrow{DA}, \overrightarrow{DE})$.\\
De plus, la somme des angles d'un triangle fait $\pi$ radians (toujours avec la même orientation). Donc on a 
\begin{eqnarray*}
&& (\overrightarrow{ED}, \overrightarrow{EA})+(\overrightarrow{DA}, \overrightarrow{DE})+(\overrightarrow{AE}, \overrightarrow{AD})=\pi \quad [2\pi]\\
 &\iff& 2(\overrightarrow{ED}, \overrightarrow{EA})+(\overrightarrow{AE}, \overrightarrow{AD})=\pi \quad [2\pi]\\
 &\iff& 2(\overrightarrow{ED}, \overrightarrow{EA})=\pi -(\overrightarrow{AE}, \overrightarrow{AD})\quad [2\pi]\end{eqnarray*}
Or $(\overrightarrow{AE}, \overrightarrow{AD})
= (\overrightarrow{AB}, \overrightarrow{AD})- (\overrightarrow{AB}, \overrightarrow{AE})=\dfrac{\pi}{2}-\dfrac{\pi}{3}=\dfrac{\pi}6 \quad [2\pi]$\\
On en déduit que $2(\overrightarrow{ED}, \overrightarrow{EA})=\pi - \dfrac{\pi}6 = \dfrac{5\pi}{6} \quad [2\pi]$.\\
Et donc que $(\overrightarrow{ED}, \overrightarrow{EA})=\dfrac{5\pi}{12} \quad [2\pi]$
\end{enumerate}
\item %D\'eterminer une mesure de $ (\overrightarrow{BE}, \overrightarrow{BF})$ et en d\'eduire une mesure de $ (\overrightarrow{EB}, \overrightarrow{EF})$
$(\overrightarrow{BE}, \overrightarrow{BF})=(\overrightarrow{BE}, \overrightarrow{BC})+ (\overrightarrow{BC}, \overrightarrow{BF})=-\dfrac{\pi}{6}-\dfrac{\pi}{3}=-\dfrac{3\pi}{6}=-\dfrac{\pi}{2} \quad [2\pi]$\\
Ainsi le triangle $EBF$ est rectangle en $B$. De plus il est isocèle car $EB=AB=BC=BF$.\\
On sait alors que $(\overrightarrow{EB}, \overrightarrow{EF})=\dfrac{\pi}{4} \quad [2\pi]$
\item \begin{enumerate}
\item %Utiliser la relation de Chasles pour calculer une mesure de $ (\overrightarrow{ED}, \overrightarrow{EF})$
$$(\overrightarrow{ED}, \overrightarrow{EF})=(\overrightarrow{ED}, \overrightarrow{EA})+(\overrightarrow{EA}, \overrightarrow{EB})+(\overrightarrow{EB}, \overrightarrow{EF})=\dfrac{5\pi}{12}+\dfrac{\pi}{3}+\dfrac{\pi}4 = \pi \quad [2\pi]$$
\item %Que peut-on en déduire sur les points $D$ , $E$ et $F$ ?
On en déduit alors que l'angle $(\overrightarrow{ED}, \overrightarrow{EF})$ est plat, donc les points $D$ $E$ et $F$ sont alignés.
\end{enumerate}
\end{enumerate}
\end{ExoF}


\end{document}
