\documentclass[mathserif,t]{beamer}
%\documentclass[mathserif, handout,t]{beamer}

\input{/home/janine/Dropbox/Travail_S/Divers/Preambule/Preambule_Cours}


\def\mySample{100}
\Teachtrue



\begin{document}

\titreCours{8}{Trigonométrie II}{}{}


 
 \begin{frame}{3. Le cercle trigonométrique}
 
Pour représenter les fonctions trigonométriques, on utilise un cercle de rayon 1, appelé le \bl{cercle trigonométrique}. 
 
 \begin{center}
 
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};
 
 \def\qmax{1};
 \def\qmin{-1};
 
 \def\a{40}
 
 % axes
 \draw[->, dblue] (0:\min) -- (0:\max) node[right]{\only<-10>{$x$}};
 \draw[->, dgreen] (90:\min) -- (90:\max) node[above]{\only<-11>{$y$}};
 
 \hunc{11}{\draw[->, dblue] (0:\min) -- (0:\max) node[right]{$x$\only<11->{= $\cos(\alpha)$}};
 \draw[->, dgreen] (90:\min) -- (90:\max) node[above]{$y$\only<12->{= $\sin(\alpha)$}};}
 
 % quadrillage
 
 \foreach \x in {\qmin, ..., \qmax}{
 \draw[black!50, very thin] (\x, \min) -- (\x, \max);}
 
 \foreach \y in {\qmin, ..., \qmax}{
 \draw[black!50,very thin]  (\min , \y) -- (\max, \y );}
 

 % origine
 \node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
\draw[dorange, thick] (0,0) circle (1cm);
 \draw[very thick] (0,0) -- node[above]{\only<4->{1}}({cos(\a)},{sin(\a)});
 \draw[->, very thick] (0:0.2) arc (0:\a:0.2) ;
 \node[right] at ({\a/2}:0.22) {$\alpha$};
 \fill ({cos(\a)}, {sin(\a)}) circle (1pt) node[right]{$P$};
 
 \hunc{2}{
\uncover<2->{
 \node at (1.2, 1.2) {I};
 \node at (-1.2, 1.2) {II};
 \node at (-1.2, -1.2) {III};
 \node at (1.2, -1.2) {IV};}
 \uncover<3->{
 \draw[very thick,dblue] (0,0) -- node[xshift=0.1cm,below]{\only<5->{$x$}} ({cos(\a)}, 0);
 \draw[very thick,dgreen] ({cos(\a)}, 0) -- node[right]{$\only<6->{y}$}({cos(\a)},{sin(\a)});
 \uncover<6->{
 \draw[very thick,dgreen] (0, 0) -- node[left]{$y$}(0,{sin(\a)});
 \draw[very thick,dgreen, dashed] ({cos(\a)}, {sin(\a)}) -- (0,{sin(\a)});
 }
}
 
 %Tangente
 \uncover<13->{
 \draw (1, \min) -- (1, \max);
 \draw[very thick, dashed] (0,0) -- (1, {tan(\a)}) node[right]{T}; 
 \draw[dblue, very thick, dashed] (0,0) -- (1, 0); 
\uncover<13->{\draw[dred, very thick] (1,0) -- node[right]{$z$} (1, {tan(\a)});} 
\uncover<16->{\draw[->, dred] (1, \min) -- (1, \max) node[above]{$\tan(\alpha)$};} 

}}
 
 \end{tikzpicture}
 
\end{center}

\hunc{7}{
\begin{multicols}{3}
\bl{$\cos(\alpha) = \dfrac{x}{1} \uncover<8->{= x}$}\\
\uncover<9->{ \ve{$\sin(\alpha) = \dfrac{y}{1} \uncover<10->{= y}$}}\\
\uncover<14->{ \ro{$\tan(\alpha) = \dfrac{z}{1} \uncover<15->{= z}$}}\\
\end{multicols}}
 
\end{frame}

\begin{frame}

\underline{Exemple 3.1} Placer les angles $50\degree$, $145\degree$, $235\degree$ et $340\degree$ sur le cercle ci-dessous et en déduire une valeur approchée de leur sinus, cosinus et tangente.  

 \begin{minipage}{0.65\textwidth}
 
 \begin{tikzpicture}[scale=2.2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\xmax{1.3};
 \def\xmin{-1.3};
 \def\ymax{1.5};
 \def\ymin{-1.3};


\draw[gray] (\xmin, \ymin) grid[step=0.1cm] ({\xmax-0.01},{\ymax-0.01});
\draw[->, thick] (1,\ymin) -- (1,\ymax) node[above]{$\tan(\alpha)$};
\draw[->, thick] (0:\xmin) -- (0:\xmax) node[below, rotate=90]{$\cos(\alpha)$};
\draw[->, thick] (90:\ymin) -- (90:\ymax) node[above]{$\sin(\alpha)$};
 
\draw[thick] (0,0) circle (1cm);
 \node[below left] at (O) {O};
 \node[above right] at (0,1) {1};
 \node[above right] at (1,0) {1};
 \node[above right] at (1,1) {1};
 \draw[thick] (-0.05, 1) -- (0.05, 1); 
 \draw[thick] (1,-0.05) -- (1, 0.05); 
 \draw[thick] (0.95,1) -- (1.05,1); 
 \fill (0,0) circle (0.5pt); 
 
 \node at (1.3, 1.6) {I};
 \node at (-1.3, 1.6) {II};
 \node at (-1.3, -1.3) {III};
 \node at (1.3, -1.3) {IV};

\hunc{2}{

\def\myAlpha{50}
\def\myRadius{0.5}

\pgfmathparse{cos(\myAlpha) }\let\ca\pgfmathresult
\draw[dred, thick, ->] (0:\myRadius) arc (0:\myAlpha:\myRadius) node[right] {\myAlpha$\degree$};
\draw[dred, thick] (0:0) -- (1, {tan(\myAlpha)}) node[right]{\only<3->{\myPrint}};
\uncover<4->{\draw[dred, dashed, thick] ({cos(\myAlpha)}, 0) node[below]{\pgfmathprintnumber{\ca}} -- ({cos(\myAlpha)},{sin(\myAlpha)});}
\uncover<5->{\draw[dred, dashed, thick] (0,{sin(\myAlpha)}) node[left]{\myPrint} -- ({cos(\myAlpha)},{sin(\myAlpha)});}

\uncover<9->{
\def\myAlpha{145}
\def\myRadius{0.4}

\pgfmathparse{cos(\myAlpha) }\let\ca\pgfmathresult
\draw[dblue, thick, ->] (0:\myRadius) arc (0:\myAlpha:\myRadius) node[above=0.2cm] {\myAlpha$\degree$};
\draw[dblue, thick] ({cos(\myAlpha)},{sin(\myAlpha)}) -- (1, {tan(\myAlpha)}) node[right]{\only<10->{\myPrint}};
\uncover<11->{\draw[dblue, dashed, thick] ({cos(\myAlpha)}, 0) node[below]{\pgfmathprintnumber{\ca}} -- ({cos(\myAlpha)},{sin(\myAlpha)});}
\uncover<12->{\draw[dblue, dashed, thick] (0,{sin(\myAlpha)}) node[right]{\myPrint} -- ({cos(\myAlpha)},{sin(\myAlpha)});}}


\uncover<16->{
\def\myAlpha{235}
\def\myRadius{0.3}

\pgfmathparse{cos(\myAlpha) }\let\ca\pgfmathresult
\draw[dgreen, thick, ->] (0:\myRadius) arc (0:\myAlpha:\myRadius) node[left] {\myAlpha$\degree$};
\draw[dgreen, thick] ({cos(\myAlpha)},{sin(\myAlpha)}) -- (1, {tan(\myAlpha)}) node[right]{\only<17->{\myPrint}};
\uncover<18->{\draw[dgreen, dashed, thick] ({cos(\myAlpha)}, 0) node[above]{\pgfmathprintnumber{\ca}} -- ({cos(\myAlpha)},{sin(\myAlpha)});}
\uncover<19->{\draw[dgreen, dashed, thick] (0,{sin(\myAlpha)}) node[right]{\myPrint} -- ({cos(\myAlpha)},{sin(\myAlpha)});}}

\uncover<23->{
\def\myAlpha{340}
\def\myRadius{0.2}

\pgfmathparse{cos(\myAlpha) }\let\ca\pgfmathresult
\draw[dorange, thick, ->] (0:\myRadius) arc (0:\myAlpha:\myRadius) node[below=0.2cm] {\myAlpha$\degree$};
\draw[dorange, thick] (0:0) -- (1, {tan(\myAlpha)}) node[right]{\only<24->{\myPrint}};
\uncover<25->{\draw[dorange, dashed, thick] ({cos(\myAlpha)}, 0) node[above left, xshift=0.1cm]{\pgfmathprintnumber{\ca}} -- ({cos(\myAlpha)},{sin(\myAlpha)});}
\uncover<26->{\draw[dorange, dashed, thick] (0,{sin(\myAlpha)}) node[left]{\myPrint} -- ({cos(\myAlpha)},{sin(\myAlpha)});}}


}
 
 \end{tikzpicture}
 
\end{minipage}\begin{minipage}{0.34\textwidth}

\hunc{2}{
\ro{
\uncover<6->{$\cos(50\degree)= 0.64$}\\
\uncover<7->{$\sin(50\degree)= 0.77$}\\
\uncover<8->{$\tan(50\degree)= 1.19$}\\}

\bl{
\uncover<13->{$\cos(145\degree)= -0.82$}\\
\uncover<14->{$\sin(145\degree)= 0.57$}\\
\uncover<15->{$\tan(145\degree)= -0.7$}\\}

\ve{
\uncover<20->{$\cos(235\degree)= -0.57$}\\
\uncover<21->{$\sin(235\degree)= -0.82$}\\
\uncover<22->{$\tan(235\degree)= 1.43$}\\}

\ora{
\uncover<27->{$\cos(340\degree)= 0.94$}\\
\uncover<28->{$\sin(340\degree)= -0.34$}\\
\uncover<29->{$\tan(340\degree)= -0.36$}\\}


}

\end{minipage}



\end{frame}

 
\begin{frame}

\underline{Définition 3.1} Le \bl{cercle trigonométrique} est composé de:
\begin{enumerate}
\item un système d'axes
\item<2-> un cercle de rayon 1 centré à l'origine 
\item<3-> un axe vertical passant par le point (1,0) 
\end{enumerate}
\uncover<4->{Le cercle trigonométrique nous permet de \bl{généraliser les fonctions trigonométriques $\sin(\alpha)$, $\cos(\alpha)$ et $\tan(\alpha)$ aux angles non aigus}. }

\vspace{0.2cm}
\uncover<5->{
\underline{Exemple 3.1} Calculer avec la calculatrice les expressions suivantes. Résoudre ensuite l'équation $\sin(x)=0.5$ pour les angles entre $0\degree$ et $360\degree$. Que remarque-t-on?
\hunc{6}{
\begin{multicols}{2}
\begin{enumerate}
\item<6-> $\sin(30\degree) \uncover<7->{= 0.5}$   
\item<8-> $\sin(150\degree) \uncover<9->{= 0.5}$      
\end{enumerate}
\end{multicols}
\uncover<10->{On résoud l'équation  $$\sin(x)=0.5 
\uncover<11->{\Rightarrow x = \sin^{-1}(0.5)}
\uncover<12->{\Leftrightarrow x = 30\degree}$$}
\uncover<13->{L'équation $\sin(x)=0.5$ a donc plusieurs solutions ($30\degree$ et $150\degree$), mais la calculatrice n'en donne qu'une seule ($x=30\degree$).\uncover<14->{ On doit utiliser le cercle trigonométrique pour trouver les autres solutions.}}
}}

\end{frame} 
  
\begin{frame}

\underline{Exercice 3.1} Pour quels angles entre $0\degree$ et $360\degree$ a-t-on $\sin(\alpha)=0.8$?\\
 %
 
 \begin{minipage}{0.54\textwidth}
 
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};
 
 \def\qmax{1};
 \def\qmin{-1};
 
 \def\a{53.13}

 \draw[->, dblue, thick] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, dgreen, thick] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, dred, thick] (1,\min) -- (1,\max) node[above]{$\tan(\alpha)$};

\draw[gray] (\min,\min) grid[step=0.1cm] ({\max-0.01},{\max-0.01});

 \node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[dorange, thick] (0,0) circle (1cm);
 
 \pgfmathparse{cos(\a) }\let\ca\pgfmathresult
 \pgfmathparse{sin(\a) }\let\sa\pgfmathresult
 \pgfmathparse{tan(\a) }\let\ta\pgfmathresult
 \pgfmathparse{cos(-\a) }\let\mca\pgfmathresult
 \pgfmathparse{sin(-\a) }\let\msa\pgfmathresult
 \pgfmathparse{tan(-\a) }\let\mta\pgfmathresult
 %
 

 \hunc{2}{ 
 \uncover<2->{\fill (0,\sa) circle (0.5pt) node[above left]{0.8};}


 \uncover<3->{\draw[dashed, dblue] (0,\sa) -- (\ca,\sa);}
 \uncover<4->{\draw[thick, dblue] (0,0) -- (\ca,\sa);}
 \uncover<4->{\fill[dblue] (\ca,\sa) circle (0.5pt);}
 \uncover<5->{\draw[thick, ->, dblue] (0:0.5) arc (0:\a:0.5);}
 \uncover<6->{\draw[thick, dblue] (\a/2:0.5) node[right] {$57.13\degree$};}
 
 \uncover<7->{\draw[dashed, dred] (0,\sa) -- (-\ca,\sa);}
 \uncover<8->{\draw[thick, dred] (0,0) -- (-\ca,\sa);}
 \uncover<8->{\fill[dblue] (-\ca,\sa) circle (0.5pt);}
 \uncover<9->{\draw[thick, dgreen] (0:-0.5) arc (0:{-\a}:-0.5);}
 \uncover<9->{\draw[thick, dgreen] (150:0.5) node[left] {$53.13\degree$};}
 \uncover<10->{\draw[thick, ->, dred] (0:0.25) arc (0:{180-\a}:0.25);}
 \uncover<11->{\draw[thick, dred] (90:0.35) node {$126.87\degree$};}}
 
 
 \end{tikzpicture}
 
 \end{minipage} \begin{minipage}{0.44\textwidth}
 

 \hunc{12}{
 \uncover<12->{
 On trouve deux solutions: 
 
$$\bl{\alpha_1= 53.13\degree}$$
\uncover<13->{ $$\ro{\alpha_2= \ve{180 - }\bl{53.13} = 126.87\degree}$$}
 }}
 
 \end{minipage}
 
 \vfill
 \uncover<14->{
 \underline{Règle 3.1} On peut observer que 
 $$\boxed{\sin(\bl{\alpha}) = \sin(\ve{180-}\bl{\alpha})}$$}
 %
 \end{frame}
 

 \begin{frame}
 
 \underline{Exercice 3.2} Pour quels angles entre $0\degree$ et $360\degree$ a-t-on $\cos(\alpha)=0.3$?\\
 
 \begin{minipage}{0.54\textwidth}
 
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};
 
 \def\qmax{1};
 \def\qmin{-1};
 
 \def\a{72.54}
 
 % axes
 \draw[->, dblue, thick] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, dgreen, thick] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, dred, thick] (1,\min) -- (1,\max) node[above]{$\tan(\alpha)$};
 
 % quadrillage
\draw[gray] (\min,\min) grid[step=0.1cm] ({\max-0.01},{\max-0.01});
 % origine
 \node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[dorange, thick] (0,0) circle (1cm);
 
 \pgfmathparse{cos(\a) }\let\ca\pgfmathresult
 \pgfmathparse{sin(\a) }\let\sa\pgfmathresult
 \pgfmathparse{tan(\a) }\let\ta\pgfmathresult
 \pgfmathparse{cos(-\a) }\let\mca\pgfmathresult
 \pgfmathparse{sin(-\a) }\let\msa\pgfmathresult
 \pgfmathparse{tan(-\a) }\let\mta\pgfmathresult
 
 \hunc{2}{
 \uncover<2->{\fill (\ca, 0) circle (0.5pt) node[below]{0.3};}
 
 \uncover<3->{\draw[dashed, dblue] (\ca,0) -- (\ca,\sa);}
 \uncover<4->{\draw[thick, dblue] (0,0) -- (\ca,\sa);}
 \uncover<4->{\fill[dblue] (\ca,\sa) circle (0.5pt);}
 \uncover<5->{\draw[thick, ->, dblue] (0:0.5) arc (0:\a:0.5);}
 \uncover<6->{\draw[thick, dblue] (\a/2:0.5) node[right] {$72.54\degree$};}
 
 \uncover<7->{\draw[dashed, dred] (\ca,0) -- (\ca,-\sa);}
 \uncover<8->{\draw[thick, dred] (0,0) -- (\ca,-\sa);}
 \uncover<8->{\fill[dblue] (\ca,-\sa) circle (0.5pt);}
 \uncover<9->{\draw[thick, dgreen] (0:0.5) arc (0:-72.54:0.5);}
 \uncover<9->{\draw[thick, dgreen] (-43:0.5) node[below, xshift=0.4cm] {$72.54\degree$};}
 \uncover<10->{\draw[thick, ->, dred] (0:0.25) arc (0:287.46:0.25);}
 \uncover<11->{\draw[thick, dred] (135:0.25) node[left] {$287.46\degree$};}}
 
 \end{tikzpicture}
 
 \end{minipage} \begin{minipage}{0.44\textwidth}
  
 \hunc{11}{
 \uncover<12->{
 On trouve deux solutions: 
 
 $$\bl{\alpha_1=72.54^\circ}$$
 \uncover<13->{$$\ro{\alpha_2=\ve{360-}\bl{72.54}=287.46^\circ}$$}
 }}
 
 
 \end{minipage}

 \vfill
 \uncover<14->{
 \underline{Règle 3.2} On peut observer que 
 $$\boxed{\cos(\bl{\alpha}) = \cos(\ve{360-}\bl{\alpha})}$$}
 

 \end{frame}

\begin{frame}

 \underline{Exemple 3.3} Pour quels angles entre $0\degree$ et $360\degree$ a-t-on $\tan(\alpha)=2$?\\
 
 \vfill

\begin{minipage}{0.54\textwidth}

\begin{tikzpicture}[scale=1.7]

\scriptsize

\coordinate (O) at (0,0);

\def\xmax{1.2}\def\xmin{-1.2}\def\ymax{2.2}\def\ymin{-2.2}
\def\xqmax{1}\def\xqmin{-1}\def\yqmax{2}\def\yqmin{-2}

\def\a{63.43}
\pgfmathparse{tan(\a) }\let\ta\pgfmathresult


\draw[->, dblue, thick] (0:\xmin) -- (0:\xmax) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
\draw[->, dgreen, thick] (90:\ymin) -- (90:\ymax) node[above]{$\sin(\alpha)$};
\draw[->, dred, thick] (1,\ymin) -- (1,\ymax) node[above]{$\tan(\alpha)$};

\draw[gray] (\xmin,\ymin) grid[step=0.1cm] ({\xmax-0.01},{\ymax-0.01});
%
\node[above left] at (0,1) {1};
\node[below right] at (1,0) {1};
\node[right] at (1,1) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 
\draw[thick] (0.95,1) -- (1.05, 1); 
\fill (0,0) circle (0.5pt); 
 %
\draw[thick] (0,0) circle (1cm);
%
%\uncover<2->{\fill[dorange] (0,\ta) circle (1pt) node[left]{$2$};} 

\hunc{2}{
\uncover<3->{
%\draw[dorange, dashed, very thick] (0,\ta) -- (1,\ta); 
\fill[dorange] (1,\ta) circle (1pt) node[above right]{2};
}

\uncover<4->{\draw[dblue, very thick] (0,0) -- (1,\ta);}
\uncover<5->{
\draw[dblue,->, thick] (0:0.4) arc (0:\a:0.4);
\node[dblue,right] at ({\a/2}:0.4) {$63.43^\circ$};
}

\uncover<7->{\draw[dred, very thick] (1,\ta) -- ({cos(180+\a)}, {sin(180+\a)});}
\uncover<8->{
\draw[dred,->, thick] (0:0.2) arc (0:{180+\a}:0.2);
\node[dred,left] at (125:0.2) {$243.43^\circ$};
}}
%

\end{tikzpicture}

\end{minipage} \begin{minipage}{0.44\textwidth}

\hunc{6}{
\uncover<6->{On obtient comme solutions

$$\bl{\alpha_1= 63.43^\circ}$$

\uncover<9->{et
$$\ro{\alpha_2= \ve{180+}\bl{63.43} = 243.43^\circ}$$}

}}


\ifTeach \vfill \else \vspace{4.5cm} \fi
 \uncover<10->{
 \underline{Règle 3.3} On peut observer que 
 $$\boxed{\tan(\bl{\alpha}) = \tan(\ve{180+}\bl{\alpha})}$$}
\end{minipage}

 

\end{frame}


\begin{frame}{Angles particuliers}

\underline{Exemple 3.4} Trouver la valeur exacte du sinus, cosinus et de la tangente des angles $30\degree$ et $60\degree$.\\[0.5cm]


\begin{minipage}{0.4\textwidth} 

\begin{tikzpicture}[scale=1.5]

\draw[thick, gray] (0,0) node[left]{A} -- node[left, dblue]{2} (60:2) node[above]{B} -- (2,0) node[above]{C} -- (0,0);
\draw[thick, dashed, gray] (60:2) -- (1,0) node[below]{D};
\hunc{3}{
\draw[thick, black] (0,0) -- (60:2) -- (1,0) -- (0,0);
\uncover<4->{\draw[dblue] (0:0.15) arc (0:60:0.15); 
\node[right, dblue] at (35 :0.15) {60\degree}; }
\uncover<5->{\draw[dblue] ($(60:2)+(240:0.3)$) arc (240:270:0.3); 
\node[below,yshift=-0.1cm, dblue] at ($(60:2)+(255:0.3)$) {30\degree}; }
\uncover<6->{\path (0,0) -- node[below, dblue]{1} (1,0);}
\uncover<11->{\path (1,0) -- node[right, dblue]{$\sqrt{3}$} (60:2);}

}

\end{tikzpicture}

\end{minipage}\begin{minipage}{0.59\textwidth} 
\hunc{2}{\small
Soit ABC un triangle équilatéral de côté $2$ et de hauteur BD. Le triangle $ABD$ est rectangle. \uncover<7->{Par Pythagore, la hauteur mesure
\begin{eqnarray*}
\overline{BD} 
\uncover<8->{&=& \sqrt{\overline{AB}^2 - \overline{AD}^2} }
\uncover<9->{= \sqrt{2^2 - 1^2} }
\uncover<10->{= \sqrt{3} }
\end{eqnarray*}
}
}

\end{minipage}
\hunc{12}{
\begin{multicols}{2}
$\cos(60\degree) 
\uncover<13->{= \dfrac{1}{2}}$ \\
\uncover<14->{ $\sin(60\degree) 
\uncover<15->{= \dfrac{\sqrt{3}}{2}}$ }\\
\uncover<16->{ $\tan(60\degree) 
\uncover<17->{= \dfrac{\sqrt{3}}{1}}\uncover<18->{ = \sqrt{3}}$ }\\
\uncover<19->{$\cos(30\degree) 
\uncover<20->{= \dfrac{\sqrt{3}}{2}}$}\\
\uncover<21->{$\sin(30\degree) 
\uncover<22->{= \dfrac{1}{2}}$ } \\
\uncover<23->{$\tan(30\degree) 
\uncover<24->{= \dfrac{1}{\sqrt{3}}}$ }
\end{multicols}
}
 
\end{frame}

\begin{frame}

\underline{Exemple 3.5} Trouver la valeur exacte du sinus, cosinus et de la tangente de l'angle $45\degree$.\\[0.5cm]


\begin{minipage}{0.4\textwidth} 

\begin{tikzpicture}[scale=1.8]

\hunc{2}{
\draw[thick, gray] (0,0) node[left]{A} 
-- node[left, dblue]{1} (0,1) node[left]{B} 
-- node[above, dblue]{1} (1,1) node[right]{C} 
-- node[right, dblue]{1} (1,0) node[right]{D} 
-- node[below, dblue]{1}(0,0);
\draw[thick, dashed, gray] (0,0) -- (1,1);
\hunc{4}{
\draw[thick, dred] (0,0) -- (1,1) -- (1,0) -- (0,0);
\uncover<6->{\draw[dblue] (0:0.15) arc (0:45:0.15); 
\node[right, dblue] at (35 :0.15) {45\degree}; }
%\uncover<5->{\draw[dblue] ($(60:2)+(240:0.3)$) arc (240:270:0.3); 
%\node[below,yshift=-0.1cm, dblue] at ($(60:2)+(255:0.3)$) {30\degree}; }
%\uncover<6->{\path (0,0) -- node[below, dblue]{1} (1,0);}
\uncover<11->{\path (0,0) -- node[left, dblue]{$\sqrt{2}$} (1,1);}

}}

\end{tikzpicture}

\end{minipage}\begin{minipage}{0.59\textwidth} 
\hunc{3}{\small
Soit ABCD un carré de côté 1.\uncover<5->{ On considère le triangle rectangle ACD.}\uncover<7->{Par Pythagore, la hauteur mesure
\begin{eqnarray*}
\overline{AC} 
\uncover<8->{&=& \sqrt{\overline{AD}^2 + \overline{BD}^2} }
\uncover<9->{= \sqrt{1^2 + 1^2} }
\uncover<10->{= \sqrt{2} }
\end{eqnarray*}
}
}

\end{minipage}
\hunc{12}{
\begin{multicols}{3}
$\cos(45\degree) 
\uncover<13->{= \dfrac{1}{\sqrt{2}}}$ \\
\uncover<14->{ $\sin(45\degree) 
\uncover<15->{= \dfrac{1}{\sqrt{2}}}$ }\\
\uncover<16->{ $\tan(45\degree) 
\uncover<17->{= \dfrac{1}{1}}
\uncover<18->{= 1}$}\\
\end{multicols}
}
 
\end{frame}

\begin{frame}

\underline{Exemple 3.6} Trouver la valeur exacte de $\tan(750\degree)$ en vous aidant du cercle trigonométrique.\\[0.2cm]

\begin{center}
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};

 \draw[->, thick] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, thick] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, thick] (1,\min) -- (1,\max) node[above]{$\tan(\alpha)$};
  
 %\node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \node[right] at (1,1) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (0.95, 1) -- (1.05, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[dorange, thick] (0,0) circle (1cm);
 

 \bigangle[dblue, thick]{750}{0.1};
 \draw[dblue, thick] (0,0) -- (1,{1/sqrt(3)});
 \hunc{3}{ \fill[dblue] (1,{tan(30)}) circle (1pt);}
  \hunc{4}{
  \draw[dgreen, ->, thick] (0:0.6) arc (0:30:0.6);
  \node[right, dgreen] at (15:0.6) {$30\degree$};
  }
  
  
\end{tikzpicture}
\end{center}

\hunc{2}{
$$\tan(\bl{750}\degree)
\uncover<5->{= \tan(\ve{30}\degree + 2 \cdot 360\degree)}
\uncover<6->{= \tan(\ve{30}\degree)}
\uncover<7->{= \dfrac{1}{\sqrt{3}}}
$$

}

\end{frame}

\begin{frame}

\underline{Exemple 3.6} Trouver la valeur exacte de $\sin(120\degree)$ en vous aidant du cercle trigonométrique.\\[0.2cm]

\begin{center}
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};

 \draw[->, thick] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, thick] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, thick] (1,\min) -- (1,\max) node[above]{$\tan(\alpha)$};
  
 \node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \node[right] at (1,1) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (0.95, 1) -- (1.05, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[thick] (0,0) circle (1cm);
  
 \draw[dred, ->, thick] (0:0.2) arc (0:120:0.2);
 \draw[dred, thick] (0:0) -- (120:1);
 \node[above right,dred] at (0:0.2) {$120\degree$};
 \hunc{3}{\draw[dred, thick, dashed] (120:1) -- (0,{sin(60)});
 \fill[dred] (0,{sin(60)}) circle (1pt);
 }
 \hunc{4}{\draw[dblue, thick, dashed] (60:1) -- (0,{sin(60)});}
 \hunc{5}{ \draw[dblue, thick] (0:0) -- (60:1);}
  \hunc{6}{\draw[dgreen, <-, thick] (120:0.6) arc (120:180:0.6);
 \node[above left,dgreen] at (180:0.6) {$60\degree$};}
 \hunc{8}{ \draw[dblue, ->, thick] (0:0.6) arc (0:60:0.6); 
 \node[above right,dblue] at (0:0.65) {$60\degree$};}

  
\end{tikzpicture}
\end{center}

\hunc{2}{
$$\sin(\ro{120}\degree) 
\uncover<7->{ = \sin(180\degree-\ve{60}\degree)}
\uncover<9->{ = \sin(\bl{60}\degree)}
\uncover<10->{ = \dfrac{\sqrt{3}}{2}}
$$

}

\end{frame}

\begin{frame}

\underline{Exemple 3.6} Trouver la valeur exacte de $\cos(225\degree)$ en vous aidant du cercle trigonométrique.\\[0.2cm]

\begin{center}
 \begin{tikzpicture}[scale=2]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};

 \draw[->, thick] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, thick] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, thick] (1,\min) -- (1,\max) node[above]{$\tan(\alpha)$};
  
 \node[below right] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \node[right] at (1,1) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (0.95, 1) -- (1.05, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[thick] (0,0) circle (1cm);
  
 \draw[dred, ->, thick] (0:0.2) arc (0:225:0.2);
 \draw[dred, thick] (0:0) -- (225:1);
 \node[left,dred] at (100:0.2) {$225\degree$};
 \hunc{2}{
 \draw[dashed, dred, thick] (225:1) -- ({cos(225)},0);
 \fill[dred] ({cos(225)},0) circle (0.8pt);}
 \hunc{3}{\draw[dblue, ->, thick] (180:0.4) arc (180:225:0.4) node[left]{$45\degree$};}
 \hunc{4}{\draw[dblue, thick] (0,0) -- ({cos(225)},0);}
 \hunc{6}{\draw[dgreen, ->, thick] (0:0.4) arc (0:45:0.4) node[right]{$45\degree$};
 \draw[dgreen, dashed, thick] (45:1) -- ({cos(45)}, 0);
 \draw[dgreen, thick] (0:0) -- (45:1);}
 \hunc{7}{\draw[dorange, thick] (0,0) -- ({cos(45)},0);}
 \hunc{8}{
 \draw[violet, thick] (-0.6,-0.05) -- (-0.5, 0.05);
 \draw[violet, thick] (-0.5,-0.05) -- (-0.4, 0.05);
 \draw[violet, thick] (0.4,-0.05) -- (0.5, 0.05);
 \draw[violet, thick] (0.5,-0.05) -- (0.6, 0.05);
 } 

  
\end{tikzpicture}
\end{center}

\hunc{2}{
\begin{eqnarray*}
\cos(\ro{225}\degree)
&=&\uncover<5->{ \cos(180\degree + \bl{45}\degree)}
\uncover<9->{= -\cos(\ve{45}\degree)}
\uncover<10->{= -\dfrac{1}{\sqrt{2}}}
\end{eqnarray*}}


\end{frame}

\begin{frame}{4. Triangles quelconques}
  
 \underline{Rappel: Relations fondamentales}
 
 \begin{multicols}{2}
 \begin{enumerate}[a)]
 \item $\cos(\alpha) = \hunc{2}{\uncover<2->{ \frac{\mbox{\bl{adj}}}{\mbox{\ro{hyp}}}}}$
 \item $\sin(\alpha) = \hunc{3}{ \frac{\mbox{\ve{opp}}}{\mbox{\ro{hyp}}}}$
 \item $\tan(\alpha) = \hunc{4}{ \frac{\mbox{\ve{opp}}}{\mbox{\bl{adj}}}}$
 \end{enumerate}
 
 \uncover<5->{Si $\ro{hyp= 1}$, alors;
 \begin{enumerate}[a)]
 \item $\cos(\alpha) = \hunc{6}{ \frac{\mbox{\bl{adj}}}{\ro{1}} \uncover<7->{= \mbox{\bl{adj}}}}$
 \item $\sin(\alpha) = \hunc{9}{ \frac{\mbox{\ve{opp}}}{\ro{1}} \uncover<10->{= \mbox{\ve{opp}}}}$
 \end{enumerate}}
 \end{multicols}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}[scale=3]
 
 \draw[thick] (0,0) -- node[below, dblue]{adj\only<8->{=cos($\alpha$)}} (0.8666, 0) -- node[right, dgreen]{opp\only<11->{=sin($\alpha$)}} (0.8666, 0.5 ) -- node[above, sloped, dred]{hyp\only<5->{=1}} (0,0); 
 \draw[thick] (0:0.2) arc (0:30:0.2) node[right]{$\alpha$};
 
 \end{tikzpicture}
 
 \end{minipage} \begin{minipage}{0.54\textwidth}
 
 \hunc{12}{
 \uncover<12->{On a donc 
 $$\boxed{\tan(\alpha) = \frac{\mbox{\ve{opp}}}{\mbox{\bl{adj}}} = \uncover<13->{\frac{\ve{\sin(\alpha)}}{\bl{\cos(\alpha)}}}}$$}}
 
 \end{minipage}
 
 \vspace{0.5cm}
 \hunc{14}{
 \uncover<14->{
 Par Pythagore, on a que  \uncover<15->{\bl{adj}$^2$ + \ve{opp}$^2$ = \ro{hyp}$^2$, et donc
 
 $$\uncover<16->{\boxed{\bl{\cos(\alpha)}^2 + \ve{\sin(\alpha)}^2 = \ro{1}}}$$ 
 
 }}}
 
 \end{frame}
 

 \begin{frame}

Soit ABC un triangle quelconque de côtés a,b,c et d'angles $\alpha, \beta, \gamma$. \\[0.2cm]
 
\framebox{\parbox{\textwidth}{ 
 \begin{minipage}{0.45\textwidth}
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
 
 \coordinate (A) at (0,0);
 \coordinate (B) at (4,0);
 \coordinate (C) at (3,3);
 \coordinate (H) at (A-|C);
 \coordinate (C2) at (2,0);
 \coordinate (C3) at (2,4);
 \coordinate (B2) at (1.5,1.5);
 \coordinate (B3) at (2.5,0.5);
 \coordinate (M) at (intersection of C2--C3 and B2--B3);
 
 \draw[very thick] (A) -- node[dblue, below]{c} (B) node[dblue, right]{B} -- node[dblue, right]{a} (C) node[blue, above]{C} -- node[dblue, left]{b} (A) node[dblue, left]{A};
 
 \begin{scope}
 \clip (A) -- (B) -- (C);
 \draw[thick] (A) circle (0.5);
 \draw[thick] (B) circle (0.5);
 \draw[thick] (C) circle (0.5);
 \node[dblue] at ($(A)+(0.7,0.2)$) {$\alpha$};  
 \node[dblue] at ($(B)+(-0.6,0.4)$) {$\beta$};  
 \node[dblue] at ($(C)+(-0.2,-0.8)$) {$\gamma$};  
 \end{scope}
\hunc{2}{
 \draw[dorange, very thick] (C) -- node[left]{h} (H) node[below]{H};
 \draw[dorange] ($(H)+(-0.3,0.1)$) -- +(0.2,0) -- +(0.2,0.2);
 \uncover<3->{\draw[dgreen, very thick] (A) -- node[below]{x}(H);}
 \uncover<4->{\draw[dred, very thick] (B) -- node[below]{y}(H);}
 
}
 
 \end{tikzpicture}
 \end{center}
 \end{minipage}\begin{minipage}{0.55\textwidth}
 
 \textbf{Théorème du cosinus}
 \begin{eqnarray*}
 \bl{a}^2 &=& \ve{b}^2 + \ro{c}^2 - 2\ve{b}\ro{c}\cos(\bl{\alpha})\\
 \ve{b}^2 &=& \bl{a}^2 + \ro{c}^2 - 2\bl{a}\ro{c}\cos(\ve{\beta})\\
 \ro{c}^2 &=& \bl{a}^2 + \ve{b}^2 - 2\bl{a}\ve{b}\cos(\ro{\gamma})\\
 \end{eqnarray*}
 \end{minipage}}}

\hunc{2}{
\small
\textbf{Démonstration.} Nous démontrons la relation $\ve{b}^2 = \bl{a}^2 + \ro{c}^2 - 2\bl{a}\ro{c}\cos(\ve{\beta})$.\\ Appelons $x$ le segment $\overline{AH}$ et $y$ $\overline{HB}$. On a
$$\sin(\beta) \uncover<5->{= \dfrac{h}{a}}\uncover<6->{ \Rightarrow h=a\sin(\beta)} \qquad
\uncover<7->{\cos(\beta) }\uncover<8->{= \dfrac{y}{a}}\uncover<9->{ \Rightarrow y=a\cos(\beta)}$$
\uncover<10->{
Par Pythagore:
\begin{eqnarray*}
b^2 &=& x^2 +  h^2
 \uncover<11->{= (c - a\cos(\beta))^2 + (a\sin(\beta))^2\\}
 \uncover<12->{&=& c^2 - 2ac\cos(\beta)+ a^2\cos^2(\beta) + a^2\sin^2(\beta)\\}
 \uncover<13->{&=& c^2 - 2ac\cos(\beta)+ a^2(\cos^2(\beta) + \sin^2(\beta))\\}
 \uncover<14->{&=& c^2 - 2ac\cos(\beta)+ a^2}
 \uncover<15->{= a^2 + c^2 - 2ac\cos(\beta) \qquad \qed} 
 \end{eqnarray*}}
 
 }
 
%
  %
 
 \end{frame}
 
 
 \begin{frame}
 
 Soit ABC un triangle quelconque de côtés a,b,c et d'angles $\alpha, \beta, \gamma$. \\[0.2cm]
  \framebox{\parbox{\textwidth}{
 \begin{minipage}{0.45\textwidth}
 \begin{center}
 \begin{tikzpicture}[scale=0.9]
 
 \draw[very thick] (A) -- node[dred, below]{c} (B) node[dgreen, right]{B} -- node[dblue, right]{a} (C) node[dred, above]{C} -- node[dgreen, left]{b} (A) node[dblue, left]{A};
 
 \begin{scope}
 \clip (A) -- (B) -- (C);
 \draw[thick] (A) circle (0.5);
 \draw[thick] (B) circle (0.5);
 \draw[thick] (C) circle (0.5);
 \node[dblue] at ($(A)+(0.7,0.2)$) {$\alpha$};  
 \node[dgreen] at ($(B)+(-0.6,0.4)$) {$\beta$};  
 \node[dred] at ($(C)+(-0.2,-0.8)$) {$\gamma$};  
 \uncover<3->{
 \draw[dorange, very thick] (C) -- node[left]{h} (H) node[below]{H};
 \draw[dorange] ($(H)+(-0.3,0.1)$) -- +(0.2,0) -- +(0.2,0.2);
 }
 \end{scope}
 \end{tikzpicture}
 \end{center}
 \end{minipage}\begin{minipage}{0.55\textwidth}
 
 \uncover<2->{
 \textbf{Théorème de l'aire}
 \begin{eqnarray*}
 \mathcal{A} &=&\frac{1}{2}\bl{a}\ve{b}\sin(\ro{\gamma}) \\
 \mathcal{A} &=& \frac{1}{2}\ve{b}\ro{c}\sin(\bl{\alpha})\\
 \mathcal{A} &=& \frac{1}{2}\bl{a}\ro{c}\sin(\ve{\beta})
 \end{eqnarray*}
 où $\mathcal{A}$ est l'aire du triangle ABC. }
 
 \end{minipage}}}
 
 
 \hunc{3}{
 \small
 \vspace{0.2cm}
 \textbf{Démonstration}: Nous démontrons l'identité $\mathcal{A}=\frac{1}{2}bc\sin(\alpha)$. Les autres identités se démontrent de manière analogue. \\[0.2cm]
 
 \uncover<4->{
 Soit $\ora{h}$ la hauteur issue de $C$. Le triangle $AHC$ est rectangle, on a donc $\ora{h=b\sin(\alpha)}$.}\uncover<5->{Par la formule de l'aire d'un triangle, on a:
 $$\mathcal{A}
 \uncover<6->{=\frac{\mbox{base}\cdot\mbox{hauteur}}{2}}
 \uncover<7->{=\frac{\ro{c}\cdot \ora{h}}{2}}
 \uncover<8->{=\frac{\ro{c}\cdot \ora{b\cdot\sin(\alpha)}}{2}}
 \uncover<9->{=\frac{1}{2}\ve{b}\ro{c}\sin(\bl{\alpha})}
 $$  
 \qed}
 }
 \end{frame}
 
 
 \begin{frame}
 
 \underline{Exemple 4.1} Soit ABC un triangle tel que $\bl{a=55}$, $\ve{b=12}$ et $\ro{\gamma=120\degree}$. Résoudre ce triangle.\\[0.2cm]

 
 \hunc{2}{
 \vfill
 On calcule l'aire: 
 $$\mathcal{A}=\frac{1}{2}\cdot\bl{a}\cdot\ve{b}\cdot\sin(\ro{\gamma}) 
 \uncover<3->{=\frac{1}{2}\cdot\bl{(55)}\cdot\ve{(12)}\cdot\sin(\ro{120}) }
 \uncover<4->{=\boxed{285.79}}$$ 
 
 \uncover<5->{
 On calcule la longueur du côté $\ro{c}$ à l'aide du théorème du cosinus:
 \begin{eqnarray*}
 &&\ro{c}^2 = \bl{a}^2 + \ve{b}^2 -2\cdot\bl{a}\cdot\ve{b}\cdot\cos(\ro{\gamma})\\
 \uncover<6->{&\Rightarrow& \ro{c}^2 = \uncover<7->{\bl{(55)}^2 + \ve{(12)}^2 -2\cdot\bl{(55)}\cdot\ve{(12)}\cdot\cos(\ro{120})}}
 \uncover<8->{= 3829}\\
 \uncover<9->{&\Leftrightarrow&  \ro{c} = \pm \sqrt{3829}\uncover<10->{ = \pm 61.88}}\\
 \uncover<11->{&\Rightarrow&  \ro{c} = \boxed{\ro{61.88}} \mbox{ (Il s'agit d'une longueur, c'est positif)}}
 \end{eqnarray*}
 }
 
 \uncover<12->{
 On utilise le théorème du cosinus pour trouver l'angle $\alpha$
 \begin{equa}
 \bl{a}^2 &=& \ve{b}^2 + \ro{c}^2 - 2\ve{b}\ro{c}\cos(\bl{\alpha}) 
 \uncover<13->{& -\bl{a^2} + 2\ve{b}\ro{c}\cos(\bl{\alpha})}\\
 \uncover<14->{\LRq 2\ve{b}\ro{c}\cos(\bl{\alpha}) &=& \ve{b}^2 + \ro{c}^2 - \bl{a}^2 }
 \uncover<15->{& \div(2\ve{b}\ro{c})}\\
 \uncover<16->{\Leftrightarrow \cos(\bl{\alpha}) &=& \dfrac{\ve{b}^2 + \ro{c}^2 - \bl{a}^2}{ 2\ve{b}\ro{c}}}
 \end{equa}}
 
 }
 
 \end{frame}
 
 \begin{frame}
 
 \hspace{2cm} \\

\begin{minipage}{0.39\textwidth}
 \begin{tikzpicture}[scale=1.5]
 
 \scriptsize
 
 \coordinate (O) at (0,0);
 \def\max{1.2};
 \def\min{-1.2};
 
 \def\qmax{1};
 \def\qmin{-1};
 
 \def\a{50.33}
 
 % axes
 \draw[->, dblue] (0:\min) -- (0:\max) node[xshift=0.2cm,rotate=90]{$\cos(\alpha)$};
 \draw[->, dgreen] (90:\min) -- (90:\max) node[above]{$\sin(\alpha)$};
 \draw[->, dred] (1,\min) -- (1,\max) node[above]{$\ro{\tan(\alpha)}$};
 
 % quadrillage
  
 % origine
 \node[below left] at (O) {O};
 \node[above left] at (0,1) {1};
 \node[below right] at (1,0) {1};
 \draw[thick] (-0.1, 1) -- (0.1, 1); 
 \draw[thick] (1,-0.1) -- (1, 0.1); 
 \fill (0,0) circle (0.5pt); 
 
 \draw[dorange, thick] (0,0) circle (1cm);
 
 \pgfmathparse{cos(\a) }\let\ca\pgfmathresult
 \pgfmathparse{sin(\a) }\let\sa\pgfmathresult
 \pgfmathparse{tan(\a) }\let\ta\pgfmathresult
 \pgfmathparse{cos(-\a) }\let\mca\pgfmathresult
 \pgfmathparse{sin(-\a) }\let\msa\pgfmathresult
 \pgfmathparse{tan(-\a) }\let\mta\pgfmathresult
 
 \hunc{5}{
 \uncover<5->{\fill (\ca, 0) circle (0.5pt) node[below right, fill=white]{0.63};}
 
 \uncover<6->{\draw[black] (\ca,-\sa) -- (\ca,\sa);}
 \uncover<7->{\draw[thick] (0,0) -- (\ca,\sa);}
 \uncover<8->{\fill (\ca,\sa) circle (0.5pt);}
 \uncover<9->{\draw[thick, ->] (0:0.2) arc (0:\a:0.2);}
 \uncover<10->{\draw[thick] (0:0.2) node[above right] {$\a^\circ$};}
 
 
 \uncover<11->{\draw[thick] (0,0) -- (\ca,-\sa);}
 \uncover<12->{\fill (\ca,-\sa) circle (0.5pt);}
 \uncover<13->{\draw[thick, ->] (0:0.15) arc (0:309.67:0.15);}
 \uncover<14->{\draw[thick] (100:0.15) node[above] {$309.67^\circ$};}}
 
 \end{tikzpicture}
 \end{minipage}\begin{minipage}{0.6\textwidth}
 
 \hunc{1}{De la relation précédente, on déduit:
 \begin{eqnarray*}
 \bl{\alpha} 
&=& \uncover<2->{\cos^{-1}\left[\frac{\ve{b}^2 + \ro{c}^2 - \bl{a}^2}{ 2\ve{b}\ro{c}}\right]} \\
 \uncover<3->{&=& \cos^{-1}\left[\frac{\ve{12}^2 + \ro{61.88}^2 - \bl{55}^2}{2\cdot\ve{(12)}\cdot\ro{(61.88)}}\right] }\\
 \uncover<4->{&=& \cos^{-1}(0.63) }
 \end{eqnarray*}}
 \end{minipage}
 
\hunc{1}{  
\vspace{0.2cm}
 $\uncover<5->{\Rightarrow\quad\alpha_1 = 50.33 \uncover<15->{\mbox{ ou } \alpha_2 = 360 - 50.33 = 309.67 }}$\\
 $\uncover<16->{\Rightarrow\quad \bl{\alpha} = \boxed{\bl{50.33}^\circ}$}\\
\uncover<17->{ (Les angles d'un triangle sont plus petits que 180$^\circ$)}\\[0.2cm]

 \uncover<18->{Finalement $\ve{\beta} = 180 - \bl{\alpha} - \ro{\gamma}
 \uncover<19->{ = 180 - \bl{50.33} - \ro{120}}
 \uncover<20->{ = \boxed{\ve{9.67}^\circ}}$ }
 
 }
 \end{frame}
 

 \begin{frame}
 
 Soit ABC un triangle quelconque de côtés a,b,c et d'angles $\alpha, \beta, \gamma$ \\
 (avec $0< \bl{\alpha} <180$, $0< \ve{\beta} <180$ et $0< \ro{\gamma} <180$)\\[0.2cm]
 
 \framebox{\parbox{\textwidth}{
 \begin{minipage}{0.45\textwidth}
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
 
 \draw[very thick] (A) -- node[dred, below, xshift=0.5cm]{c} (B) node[dgreen, right]{B} -- node[dblue, right]{a} (C) node[dred, above]{C} -- node[dgreen, left]{b} (A) node[dblue, left]{A};
 \fill[dorange] (M) circle (0.05);
 
 \draw[dorange, thick] (M) circle (2.28cm) node[above right]{M};
 \draw[dorange, thick, ->] (M) -- node[left]{r} +(110:2.28);
 \hunc{5}{
 \draw[violet, thick] (M) -- node[above]{r} (A);
 \draw[violet, thick] (M) -- node[above]{r} (B);
 \draw[violet, thick] (M) -- (C2) node[below]{H};
 \begin{scope}
 \clip (A) -- (M) -- (C2);
 \hunc{8}{ \draw[violet] (M) circle (0.5);
\node[violet] at ($(M)-(0.3,0.6)$) {$\gamma$};
 \end{scope}
 \draw[violet] (A) -- node[below]{$\only<8->{\frac{c}{2}}$} (C2);
 \draw[violet] ($(C2)+(-0.3,0.1)$) -- +(0.2,0) -- +(0.2,0.2);}
 }
 

 
 \begin{scope}
 \clip (A) -- (B) -- (C);
 \draw[thick] (A) circle (0.5);
 \draw[thick] (B) circle (0.5);
 \draw[thick] (C) circle (0.5);
 \node[dblue] at ($(A)+(0.7,0.2)$) {$\alpha$};  
 \node[dgreen] at ($(B)+(-0.6,0.4)$) {$\beta$};  
 \node[dred] at ($(C)+(-0.2,-0.8)$) {$\gamma$};  
 \end{scope}
 \end{tikzpicture}
 \end{center}
 \end{minipage}\begin{minipage}{0.55\textwidth}
 
 \uncover<2->{

 \textbf{Théorème du sinus}
 $$\frac{\bl{a}}{\sin(\bl{\alpha})} 
 = \frac{\ve{b}}{\sin(\ve{\beta})}
 = \frac{\ro{c}}{\sin(\ro{\gamma})} = 2\ora{r}$$
 où $\ora{r}$ est le rayon du cercle circonscrit 
 }

 \end{minipage}}}
 
 \hunc{3}{
 \small
 \vspace{0.2cm}
\textbf{Démonstration.} Nous démontrons $\dfrac{\ro{c}}{\sin(\ro{\gamma})} = 2\ora{r}$. \\

\uncover<4->{Soit le triangle AMB isocèle en M et MH sa hauteur. }\uncover<6->{Par le théorème de l'angle au centre, $\angle AMB = 2\gamma$.}\uncover<7->{ Le triangle est isocèle, MH est donc aussi la bissectrice et la médiatrice, on a donc que $\angle AMH=\gamma$ et $\overline{AH}=\frac{c}{2}$.}\uncover<9->{ Le triangle $AMH$ est rectangle en H, on a donc 
$$\sin(\gamma) \uncover<10->{= \dfrac{\frac{c}{2}}{r}}\uncover<11->{ = \dfrac{c}{2r}}\uncover<12->{ \Rightarrow \frac{c}{\sin(\gamma)} = 2r \quad\quad \qed }$$
 
 }}
 \end{frame}
 
 \begin{frame}
 \underline{Exemple 4.2} Résoudre le triangle ABC donné par $\bl{a=18}$, $\ve{\beta=60^\circ}$ et $\ro{\gamma=40^\circ}$. \\[0.2cm]
 
 \hunc{2}{
 \uncover<2->{On commence par calculer l'angle $\bl{\alpha}$:
 $$\bl{\alpha} = 180 -\ve{\beta} - \ro{\gamma} 
 \uncover<3->{= 180 -\ve{60} - \ro{40} }
 \uncover<4->{= \boxed{\bl{80^\circ}}}$$}
 
 \uncover<5->{
 On trouve $b$ et $c$ par le théorème du sinus:
 \begin{eqnarray*}
 \frac{\bl{a}}{\sin(\bl{\alpha})} =\frac{\ve{b}}{\sin(\ve{\beta})}
 \uncover<6->{&\Leftrightarrow& \ve{b} = \frac{\bl{a}\sin(\ve{\beta})}{\sin(\bl{\alpha})} }
 \uncover<7->{ = \frac{\bl{18}\cdot\sin(\ve{60})}{\sin(\bl{80})} }
 \uncover<8->{ = \boxed{\ve{15.83}} }\\
 \uncover<9->{
 \frac{\bl{a}}{\sin(\bl{\alpha})} =\frac{\ro{c}}{\sin(\ro{\gamma})}
 \uncover<10->{&\Leftrightarrow& \ro{c} = \frac{\bl{a}\sin(\ro{\gamma})}{\sin(\bl{\alpha})} }
 \uncover<11->{ = \frac{\bl{18}\cdot\sin(\ro{40})}{\sin(\bl{80})} }
 \uncover<12->{ = \boxed{\ro{11.75}} }}
 \end{eqnarray*}}
 
 \uncover<13->{On calcule l'aire du triangle
 $$\mathcal{A} 
 = \frac{1}{2}\bl{a}\ve{b}\sin(\ro{\gamma})
 \uncover<14->{= \frac{1}{2}\cdot\bl{18}\cdot\ve{15.83}\cdot\sin(\ro{40})}
 \uncover<15->{= \boxed{91.57} }$$}}
 
 \end{frame}
 
 
 \begin{frame}
 \underline{Exemple 4.3} Résoudre le triangle ABC tel que $\bl{a=17}$, $\ve{b=18}$ et $\bl{\alpha=65^\circ}$. L'aire n'est pas demandée. \\[0.2cm]
 
 \hunc{2}{
 \uncover<2->{On utilise le théorème du sinus pour trouver $\beta$:}
 \uncover<3->{
 \begin{eqnarray*}
 \frac{\bl{a}}{\sin(\bl{\alpha})} &=& \frac{\ve{b}}{\sin(\ve{\beta})} 
 \uncover<4->{\Leftrightarrow  \sin(\ve{\beta})=\frac{\ve{b}\sin(\bl{\alpha})}{\bl{a}}}
 \uncover<5->{=\frac{\ve{18}\cdot\sin(\bl{65}^\circ)}{\bl{17}}}
 \uncover<6->{= 0.96}
 \end{eqnarray*}}
 $$\uncover<7->{\Rightarrow \begin{cases} 
 \ve{\beta_1} = \sin^{-1}(0.96) = \ve{73.66}^\circ \mbox{ et }}\\
 \uncover<8->{\ve{\beta_2} = 180^\circ - \ve{\beta_1} = 180^\circ - \ve{73.66}^\circ = \ve{106.34}^\circ}
 \end{cases}$$
 \uncover<9->{\ro{\textbf{Les deux solutions peuvent être valides!}} On résoud pour les deux cas:}\\[0.1cm]
 \uncover<10->{\underline{Cas 1}: $\boxed{\ve{\beta = 73.66}^\circ}$}\\
 \uncover<11->{$\Rightarrow \gamma = 180^\circ - \bl{\alpha}- \ve{\beta}$} 
 \uncover<12->{$= 180^\circ - \bl{65}^\circ- \ve{73.66}^\circ$} 
 \uncover<13->{$= \boxed{\ro{41.34}^\circ}$} \\
 \uncover<14->{On calcule $c$ avec le théorème du sinus: 
 \begin{eqnarray*}
 \frac{\bl{a}}{\sin(\bl{\alpha})} &=& \frac{\ro{c}}{\sin(\ro{\gamma})} 
 \uncover<15->{\Leftrightarrow  \ro{c}=\frac{\bl{a}\sin(\ro{\gamma})}{\sin(\bl{\alpha})}} 
 \uncover<16->{=\frac{\bl{(17.34)}\cdot\sin(\ro{41.34}^\circ)}{\sin(\bl{65})}} 
 \uncover<17->{= \boxed{\ro{12.64}}} 
 \end{eqnarray*}}
  }
 
 \end{frame}
 
 \begin{frame}
 
 \hunc{1}{
\underline{Cas 2}: $\boxed{\ve{\beta = 106.34}^\circ}$\\
 \uncover<2->{$\Leftrightarrow \gamma = 180^\circ - \bl{\alpha}- \ve{\beta}$} 
 \uncover<3->{$= 180^\circ - \bl{65}^\circ- \ve{106.34}^\circ$} 
 \uncover<4->{$= \boxed{\ro{8.66}^\circ}$} \\
 \uncover<5->{On calcule $c$ avec le théorème du sinus: 
 \begin{eqnarray*}
 \frac{\bl{a}}{\sin(\bl{\alpha})} &=& \frac{\ro{c}}{\sin(\ro{\gamma})} 
 \uncover<6->{\Leftrightarrow  \ro{c}=\frac{\bl{a}\sin(\ro{\gamma})}{\sin(\bl{\alpha})}} 
 \uncover<7->{=\frac{\bl{(17.34)}\cdot\sin(\ro{8.66}^\circ)}{\sin(\bl{65})}} 
 \uncover<8->{= \boxed{\ro{2.88}}} 
 \end{eqnarray*}}
 \uncover<9->{Construisons les triangles pour vérifier si les réponses sont valides!}

 \small
 
 \uncover<10->{
 \begin{minipage}{0.49\textwidth}
 Cas 1: 
 %\begin{itemize}
 %\item $\bl{a=17.34}$, $\bl{\alpha=63.04}^\circ$
 %\item $\ve{b=18.49}$, $\ve{\beta=71.89}^\circ$
 %\item $\ro{c=13.77}$, $\ro{\gamma=45.07}^\circ$
 %\end{itemize}
 \begin{center}
 \begin{tikzpicture}[scale=0.15]
 \footnotesize
 \coordinate (A) at (0,0);
 \coordinate (C) at ($(A)+(63.04:18.49)$);
 \coordinate (B1) at (2.99,0);
 \coordinate (B2) at (13.77,0);
 
 \uncover<11->{\draw[dred, thick] (A) node[dblue, left]{A} -- node[below]{c=12.64} (B2) node[dgreen, right]{B};}
 \uncover<12->{
 \draw[dgreen, thick] (A) -- node[sloped, above]{b=18} (C) node[dred, above]{C};
 \begin{scope}
 \clip (A) -- (B2) -- (C);
 \draw (A) circle (1cm); 
 \node[above right, xshift=0.2cm] (A) {$\alpha=65^\circ$} ;
 \end{scope}
 \uncover<13->{
 \draw[dblue, thick] (C) -- node[sloped, above]{\only<14->{a=17?}\;\only<15->{OK}} (B2);
 }
 }}
 
 \end{tikzpicture}
 \end{center}
 
 \uncover<16->{Le triangle est \textbf{constructible}.\\ Cette solution est donc \textbf{valide}.}
 \end{minipage}
 \begin{minipage}{0.49\textwidth}
 \uncover<17->{
 Cas 2: 
 %\begin{itemize}
 %\item $\bl{a=17.34}$, $\bl{\alpha=63.04}^\circ$
 %\item $\ve{b=18.49}$, $\ve{\beta=71.89}^\circ$
 %\item $\ro{c=2.99}$, $\ro{\gamma=8.85}^\circ$
 %\end{itemize}
 \begin{center}
 \begin{tikzpicture}[scale=0.15]
 \footnotesize
 \coordinate (A) at (0,0);
 \coordinate (C) at ($(A)+(63.04:18.49)$);
 \coordinate (B1) at (2.99,0);
 \coordinate (B2) at (13.77,0);
 
 \uncover<18->{\draw[dred, thick] (A) node[dblue, left]{A} -- node[below]{c=2.88} (B1) node[dgreen, right]{B};}
 \uncover<19->{
 \draw[dgreen, thick] (A) -- node[sloped, above]{b=18} (C) node[dred, above]{C};
 \begin{scope}
 \clip (A) -- (B1) -- (C);
 \draw (A) circle (1cm); 
 \end{scope}
 \node[above right, xshift=0.2cm] (A) {$\alpha=65^\circ$} ;
 \uncover<20->{
 \draw[dblue, thick] (C) -- node[sloped, below]{\only<9->{a=17?}\;\only<10->{OK}} (B1);
 }
 }
 
 \end{tikzpicture}
 \end{center}
 
 \uncover<21->{Le triangle est \textbf{constructible}. \\ Cette solution est donc \textbf{valide}.}
 
 
 \end{minipage}}}
 
 \end{frame}
 
 \begin{frame}
 
 \underline{Autre méthode de vérification}\\[0.1cm]
 \small
 \begin{minipage}{0.35\textwidth}
 
 \begin{tikzpicture}[scale=0.7]
 \coordinate (A) at (0,0);
 \coordinate (B) at (4,0);
 \coordinate (C) at (3,3);
 \draw[very thick] (A) -- node[dred, below]{c} (B) node[dgreen, right]{B} -- node[dblue, right]{a} (C) node[dred, above]{C} -- node[dgreen, left]{b} (A) node[dblue, left]{A};
 \begin{scope}
 \clip (A) -- (B) -- (C);
 \draw[thick] (A) circle (0.5);
 \draw[thick] (B) circle (0.5);
 \draw[thick] (C) circle (0.5);
 \node[dblue] at ($(A)+(0.7,0.2)$) {$\alpha$};  
 \node[dgreen] at ($(B)+(-0.6,0.4)$) {$\beta$};  
 \node[dred] at ($(C)+(-0.2,-0.8)$) {$\gamma$};  
 \end{scope}
 \end{tikzpicture}
 
 \end{minipage}\; \begin{minipage}{0.63\textwidth}
 
 Dans un triangle, le plus grand angle correspond au plus grand côté et le petit au plus petit côté. Par exemple, si $$\boxed{\ve{\beta} > \ro{\gamma} > \bl{\alpha} \Leftrightarrow \ve{b} > \ro{c} > \bl{a}}$$
 
 \end{minipage}
 
 \hunc{2}{
 \begin{minipage}{0.49\textwidth}
 \uncover<2->{
 Cas 1: 
 \begin{itemize}
 \item $\bl{a=17}$, $\bl{\alpha=65}^\circ$
 \item $\ve{b=18}$, $\ve{\beta=73.66}^\circ$
 \item $\ro{c=12.64}$, $\ro{\gamma=41.34}^\circ$
 \end{itemize}
 
 \uncover<3->{
 On vérifie, on a bien:
 $$\ve{\beta} > \bl{\alpha} > \ro{\gamma} \Leftrightarrow \ve{b} > \bl{a} > \ro{c} $$ }
 
 \vspace{-0.5cm}
 \uncover<4->{Le triangle est \textbf{constructible}.\\ Cette solution est donc \textbf{valide}.}}
 \end{minipage}
 \begin{minipage}{0.49\textwidth}
 \uncover<5->{
 Cas 2: 
 \begin{itemize}
 \item $\bl{a=17}$, $\bl{\alpha=65}^\circ$
 \item $\ve{b=18}$, $\ve{\beta=106.34}^\circ$
 \item $\ro{c=2.88}$, $\ro{\gamma=8.66}^\circ$
 \end{itemize}
 
 
 \uncover<6->{
 On vérifie, on a bien:
 $$\ve{\beta} > \bl{\alpha} > \ro{\gamma} \Leftrightarrow \ve{b} > \bl{a} > \ro{c} $$ }
 
 \vspace{-0.5cm}
 \uncover<7->{Le triangle est \textbf{constructible}. \\ Cette solution est donc \textbf{valide}.}
 }
 \end{minipage}}
 
 \end{frame}

\begin{frame}{}
\underline{Exemple 4.4} Résoudre le triangle tel que $\ve{\beta=30}^\circ$, $\bl{a=10}$ et $\ve{b=36}$. L'aire n'est pas demandée.\\[0.1cm]

\hunc{2}{
\small
\uncover<2->{On utilise le théorème du sinus pour trouver $\alpha$:}
\uncover<3->{
\begin{eqnarray*}
\frac{\bl{a}}{\sin(\bl{\alpha})} &=& \frac{\ve{b}}{\sin(\ve{\beta})} 
\uncover<4->{\Leftrightarrow  \sin(\bl{\alpha})=\frac{\bl{a}\sin(\ve{\beta})}{\ve{b}}}
\uncover<5->{=\frac{\bl{10}\cdot\sin(\ve{30}^\circ)}{\ve{36}}}
\uncover<6->{= 0.14}\\
\uncover<7->{\Rightarrow \bl{\alpha_1} &=& \sin^{-1}(0.14) = \bl{7.98}^\circ} 
\uncover<8->{\mbox{ et } \bl{\alpha_2} = 180^\circ - \bl{\alpha_1} = 180^\circ - \bl{7.98}^\circ = \bl{172.02}^\circ}
\end{eqnarray*} }
\uncover<9->{On résoud pour les deux cas:}\\[0.1cm]
\uncover<10->{\underline{Cas 1}: $\boxed{\bl{\alpha = 7.98}^\circ}$}\\
\uncover<11->{$\Rightarrow \gamma = 180^\circ - \bl{\alpha}- \ve{\beta}$} 
\uncover<12->{$= 180^\circ - \bl{7.89}^\circ- \ve{30}^\circ$} 
\uncover<13->{$= \boxed{\ro{142.02}^\circ}$} \\
\uncover<14->{On calcule $c$ avec le théorème du sinus: 
\begin{eqnarray*}
\frac{\bl{a}}{\sin(\bl{\alpha})} &=& \frac{\ro{c}}{\sin(\ro{\gamma})} 
\uncover<15->{\Leftrightarrow  \ro{c}=\frac{\bl{a}\sin(\ro{\gamma})}{\sin(\bl{\alpha})}} 
\uncover<16->{=\frac{\bl{10}\cdot\sin(\ro{142.02}^\circ)}{\sin(\bl{7.98})}} 
\uncover<17->{= \boxed{\ro{44.31}}} 
\end{eqnarray*}}
\uncover<18->{On vérifie $\bl{10} < \ve{36} < \ro{44.31}$ et \uncover<19->{$\bl{7.98}\degree < \ve{30}\degree < \ro{142.02}\degree$}\uncover<20->{ $\Rightarrow$ OK}}\\
\uncover<21->{\underline{Cas 2}: $\boxed{\bl{\alpha = 172.02}^\circ}$} 
\uncover<22->{$\Rightarrow \gamma $} 
\uncover<23->{$= 180^\circ - \bl{172.02}^\circ- \ve{30}^\circ$} 
\uncover<24->{$= -22.02^\circ$}
\uncover<25->{$ <0$}\\
\uncover<26->{On ne peut pas avoir d'angle négatif dans un triangle. Le deuxième cas n'est donc pas possible.}}
\end{frame}


\begin{frame}
\def\b{7.5}\def\al{60}
\underline{Exercice de groupe} Soit un triangle tel que $\bl{\alpha= \al}\degree$ et $\ve{b= \b}$ cm. Discuter le nombre de cas possibles en fonction de la longeur du segment $a$. 

\hunc{2}{
\begin{center}
\begin{tikzpicture}[scale=3]

\draw[dashed, thick] (-0.5,0)  -- (2.5,0);
\hunc{3}{\draw[thick] (0:0)node[left]{A} -- node[above, sloped]{b=7.5 cm} (60:2) node[above]{C};
\draw[ ->, thick] (0:0.2) arc (0:60: 0.2) node[right]{$\al\degree$};}

\small
\hunc{15}{\node[right, dred] at (2.5,1.4) {$a < bsin(\alpha)$};}
\hunc{15}{\node[right, dred] at (2.5,1.3) {O sol};}
\hunc{14}{\node[right, dblue] at (2.5,1.1) {$a = bsin(\alpha)$};}
\hunc{14}{\node[right, dblue] at (2.5,1.0) {1 triangle rect.};}
\hunc{14}{\node[right, dgreen] at (2.5,0.8) {$bsin(\alpha) < a < b$};}
\hunc{14}{\node[right, dgreen] at (2.5,0.7) {2 sol.};}
\hunc{15}{\node[right, violet] at (2.5,0.5) {$a = b$: 1 sol.};}
\hunc{15}{\node[right, violet] at (2.5,0.4) {et 1 triangle plat };}
\hunc{15}{\node[right, dorange] at (2.5,0.2) {$a > b$: 1 sol.};}
\hunc{15}{\node[right, dorange] at (2.5,0.1) {et un triangle };}
\hunc{15}{\node[right, dorange] at (2.5,0) {non valide};}

\clip (-0.5,-0.5) rectangle (2.5,2);
\hunc{4}{\fill[gray, thick] (60:2) circle (0.02cm);}

%\draw[gray, ->, thick] (60:2) -- ($(60:2)+(330:1.6)$);
\hunc{5}{\draw[dred, dashed, thick] (60:2) circle (1.6cm);}

%\draw[dorange, ->, thick] (60:2) -- ($(60:2)+(325:1.73)$);
\hunc{7}{\draw[dblue, thick] (60:2) -- (1,0);
\draw[dblue, thick] (0,0) -- (1,0);}
\hunc{6}{\draw[dblue, dashed, thick] (60:2) circle (1.73cm);}

%\draw[dgreen, ->, thick] (60:2) -- ($(60:2)+(320:1.85)$);
\hunc{9}{\draw[dgreen, thick] (60:2) -- (1.65,0);
\draw[dgreen, thick] (0,0) -- (1.65,0);}
\hunc{10}{\draw[dgreen, thick] (60:2) -- (0.35,0);}
\hunc{8}{\draw[dgreen, dashed, thick] (60:2) circle (1.85cm);}

%\draw[violet, ->, thick] (60:2) -- ($(60:2)+(315:2)$);
\hunc{12}{\draw[violet, thick] (60:2) -- (2,0);
\draw[violet, thick] (0,0) -- (2,0);}
\hunc{11}{\draw[violet, dashed, thick] (60:2) circle (2cm);}

%\draw[dblue, ->, thick] (60:2) -- ($(60:2)+(310:2.1)$);
\hunc{14}{\draw[dorange, thick] (60:2) -- (2.19,0);
\draw[dorange, thick] (0,0) -- (2.19,0);}
\hunc{15}{\draw[dorange, thick, dashed] (60:2) -- (-0.19,0);
\draw[dorange, thick, dashed] (0,0) -- (-0.19,0);}
\hunc{13}{\draw[dorange, dashed, thick] (60:2) circle (2.1cm);}

\end{tikzpicture}
\end{center}}

\end{frame}

\end{document}




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