\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 I}{}{}


 \begin{frame}{1. Mesure d'angles}
 
 \uncover<2->{Un \bl{angle} est une grandeur permettant de décrire l'\bl{amplitude d'une rotation}.}\\[0.2cm]
 
 \uncover<3->{
 \begin{center}
 \begin{tikzpicture}
 \draw (0:0) -- (0:3);
 \draw (0:0) -- (30:3);
 \draw[->,dblue, thick] (0:0.5) arc (0:30:0.5) node[right] {$\alpha = 30^{\circ}$};
 \hunc{2}{
 \uncover<4->{\draw[->,dblue, thick] (0:3.2) arc (0:30:3.2);
 \node[right, dblue] at (15:3.2) {$+$};}
 \uncover<6->{\draw[->,dred, thick] (0:1.5) arc (0:-330:1.5);
 \node[right,dred] at (-1.5, 0) {$-330^{\circ}$};}
 \uncover<7->{\bigangle[dgreen, thick]{390}{0.8};}
 \uncover<11->{\bigangle[dorange, thick]{750}{1};}}
 \end{tikzpicture}
 \end{center}}
 \ifTeach \else \vspace{1.8cm} \fi
 
 \uncover<4->{On mesure les angles dans le \ve{sens contraire des aiguilles} d'une montre. }\uncover<5->{Un tour complet mesure \bl{$360^{\circ}$}.}\\[0.1cm]

\uncover<8->{Un rotation de $\bl{30^{\circ}}$ est équivalente à une rotation de $\bl{30^{\circ}}$ plus ou moins des tours complets, par exemple $30 + 360 = \ve{390^{\circ}}$ ou $30 -360 = \ro{-330^{\circ}}$.} \uncover<9->{Ce sont trois mesures du \bl{même angle}.}\\[0.1cm]

\uncover<10->{Plus généralement, l'expression $\bl{\alpha + 360\cdot k}$ avec $k\in\mathbb{Z}$ dénote toutes des mesures possibles de l'angle $\bl{\alpha}$. } 
 
 \end{frame}
 
 \begin{frame}
 
 \underline{Exemple 1.1} Donner toutes les mesures de l'angle $\bl{\alpha=50^{\circ}}$ entre -1000$\degree$ et 1000$\degree$. \\[0.5cm]

\hunc{2}{On a: 
\begin{enumerate}
\item<2-> $\bl{50}\degree$
\item<3-> $\bl{50}+\ro{1}\cdot \ve{360}=\bl{410}\degree$
\item<4-> $\bl{50}+\ro{2}\cdot \ve{360}=50+720=\bl{770}\degree$
\item<5-> $\bl{50}-\ro{1}\cdot \ve{360}=\bl{-310}\degree$
\item<6-> $\bl{50} - \ro{2}\cdot \ve{360} = 50 -720 = \bl{-670}\degree$ \\[0.5cm]
\end{enumerate} }

\ifTeach \else \vspace{4.5cm} \fi

\uncover<7->{\underline{Remarque 1.1} Pour dénoter \textbf{toutes les mesures possibles} de l'angle $\alpha= \bl{50^{\circ}}$, on écrira:
$$\alpha = \uncover<8->{\bl{50} + \ro{k}\cdot \ve{360}^{\circ}}\uncover<9->{, \mbox{ avec } \ro{k}\in\mathbb{Z}}$$  
\uncover<10->{où $\ro{k}\in\mathbb{Z}$ signifie que $\ro{k}$ est un nombre entier positif ou négatif.} }  

\end{frame}
 
 \begin{frame}{Radians}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}[scale=0.8]
 
 \def\r{2}
 \def\a{57.296}
 
 %\uncover<6->{
 %\draw[dorange, thick] (0:\r) arc (0:2*\a:\r);
 %\node[above, dorange] at (45: \r+0.15) {$L_{\beta}$=2\ro{r}};
 %\uncover<7->{\draw[dorange, thick] (0:0.3) arc (0:2*\a:0.3) node[right, yshift=0.5cm] {$\beta \only<5->{= 2 \mbox{ rad}}$};}
 %\draw[dorange] (0,0) -- (2*\a:\r);
 %}
 %
 %\uncover<9->{
 %\draw[violet, thick] (0:\r) arc (0:3*\a:\r);
 %\node[above, violet] at (90: \r+0.15) {$L_{\gamma}$=3\ro{r}};
 %\uncover<10->{\draw[violet, thick] (0:0.6) arc (0:3*\a:0.6) node[right, yshift=1.2cm, xshift=-0.2cm] {$\gamma \only<5->{= 3 \mbox{ rad}}$};}
 %\draw[violet] (0,0) -- (3*\a:\r);
 %\draw[dashed] (0,0) -- (180:\r);
 %}
 
 
 \draw (0,0) circle (\r);
 \draw (0,0) -- (\r,0);
 \draw[blue] (0,0) -- (\a:\r);
 
 \draw[dblue, thick] (0:\r) node[right, yshift=1cm, xshift=-0.25cm] {$L_{\alpha}$ \only<4->{= \ro{r}}} arc (0:\a:\r);
 \draw[dgreen, thick] (0:0.2) arc (0:\a:0.2) node[right, yshift=0.1cm] {$\alpha \only<5->{= 1 \mbox{ rad}}$};
 \draw[dred, thick] (0:0) -- node[below] {r}  (0:\r) ;
 
 \end{tikzpicture}
 
 \end{minipage} \begin{minipage}{0.54\textwidth}
 
 \uncover<2->{On appelle la longueur \bl{$L_{\alpha}$} la \bl{longueur d'arc} interceptée par l'\ve{angle $\alpha$}. Cette longueur dépend du \ro{rayon du cercle r}. }\\
 
 \uncover<3->{
 On dit qu'un angle vaut \ve{1 radian} [rad] si la \bl{longeur d'arc} qu'il intercepte est égale au \ro{rayon du cercle}:\\
 \uncover<4->{$$\boxed{\ve{\alpha = 1 \mbox{ rad}} \Leftrightarrow \bl{L_{\alpha}} = \ro{r}}$$}
}
 
 \end{minipage}
 
 
 \vspace{0.2cm}
 \uncover<8->{
 Si la \bl{longueur de l'arc} vaut \bl{$L_{\alpha}$} et le rayon \ro{$r$}, l'\ve{angle correspondant $\alpha$} sera donné par:
 
 $$\boxed{\ve{\alpha} = \frac{\bl{L_{\alpha}}}{\ro{r}}}$$}
 
 \end{frame}
 
 \begin{frame}{Passage de radians en degrés}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}
 
 \def\r{2}
 \def\a{57.296}
 
 \draw (0,0) circle (\r);
 \fill (0,0) circle (0.05);
 
 \hunc{3}{
 \uncover<3->{
 \draw[dblue] (0,0) circle (\r);
 \node[above, dblue] at (0, 2.1cm) {$L_{\alpha}=2\pi r$};
 }
 
 \uncover<4->{
 \draw[dgreen] (0,0) circle (0.2) node[right, xshift=0.2cm] {$\alpha$};
 }}
 
 \end{tikzpicture}
 
 \end{minipage} \begin{minipage}{0.54\textwidth}
 \hunc{1}{
 La circonférence d'un cercle vaut \uncover<2->{\bl{$L =2 \pi r$}}\\
 
 \uncover<5->{L'angle $\alpha$ vaut donc:
 $$\ve{\alpha} = \frac{\bl{L_{\alpha}}}{\ro{r}} \hunc{6}{\uncover<6->{ = \frac{\bl{2\pi r}}{\ro{r}}}\uncover<7->{ = \ve{2\pi} \mbox{ rad}}}$$
 }
 
 \uncover<8->{ En degrés, cet angle vaut: \uncover<9->{$\ve{\alpha = 360^{\circ}}$.}}\\}
 
 \end{minipage}
 
 \hunc{10}{
 \uncover<10->{
 On a donc:
 \begin{eqnarray*}
 &2 \pi \mbox{ rad} &\leftrightarrow 360^{\circ}\\
 \uncover<11->{\Leftrightarrow & \pi \mbox{ rad} &\leftrightarrow 180^{\circ}}\\
 \uncover<12->{\Leftrightarrow & 1 \mbox{ rad} & }
 \uncover<13->{\leftrightarrow \left(\frac{180}{\pi}\right)^{\circ}}
 \uncover<14->{\approx 57.3^{\circ} }
 \end{eqnarray*}}}
 
 \end{frame}
  
 \begin{frame}
 
 \underline{Exemple 1.1} Combien valent 5 radians en degré?\\[0.2cm]
 \hunc{2}{
 \uncover<2->{
 \begin{minipage}{0.4\textwidth}
 \begin{tikzpicture}
	 \matrix (A) [matrix of math nodes, column sep = 2mm, ampersand replacement=\&] 
	 {
		 \mbox{radians} \& \mbox{degrés}\\
		 \pi \& 180 \\
		  5 \& \only<5->{286.45} \\
	 };
			 
	 \onslide<3->{\draw[->, dgreen, thick] (A-3-1) to (A-2-2);}
	 \onslide<4->{\draw[->, dred, thick] (A-2-2) to (A-2-1);}
	 \onslide<5->{\draw[->, dblue, thick] (A-2-1) to (A-3-2);}
 
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.58\textwidth}
 \uncover<3->{ $\dfrac{\ve{5 \cdot 180}}{\uncover<4->{\ro{\pi}} }\uncover<5->{\, \bl{= 286.45^{\circ}}}$}	
 \end{minipage}}
 
 \vspace{0.2cm}
 
 \uncover<6->{\vi{ 5 radians équivalent à 286.45 degrés.} }\\}
 
 \uncover<7->{
 \vfill
 \underline{Exemple 1.2} Combien valent 30 degrés en radians?\\[0.2cm]
 
 \hunc{8}{
 \begin{minipage}{0.4\textwidth}
 \begin{tikzpicture}
	 \matrix (A) [matrix of math nodes, column sep = 2mm, ampersand replacement=\&] 
	 {
		 \mbox{radians} \& \mbox{degrés}\\
		 \pi \& 180 \\
		 \only<11->{\dfrac{\pi}{6}} \&  30 \\
	 };
			 
	 \onslide<9->{\draw[->, dgreen, thick] (A-3-2) to (A-2-1);}
	 \onslide<10->{\draw[->, dred, thick] (A-2-1) to (A-2-2);}
	 \onslide<11->{\draw[->, dblue, thick] (A-2-2) to (A-3-1);}
 
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.58\textwidth}
 \uncover<9->{ $\dfrac{\ve{30 \cdot \pi}}{ \uncover<10->{\ro{180}}} 
 \uncover<11->{=\bl{\, \dfrac{\pi}{6} \mbox{ radians}}}$}	
 \end{minipage}}}
 
 \vspace{0.2cm}
 
\hunc{12}{\vi{30 degrés équivalent à $\dfrac{\pi}{6}$ radians.}}
 
 \end{frame}
 
 \begin{frame}
 
\underline{Exemple 1.3} Placer les angles 0$\degree$, 180$\degree$, 90$\degree$, 270$\degree$, 45$\degree$, 30$\degree$ et 60$\degree$ sur le cercle suivant. Calculer la valeur de ces angles en radians.    
 
\begin{center}
\begin{tikzpicture}[scale=1.5]
 
 \def\r{2}
 \def\a{57.296}
 
 \draw (0,0) circle (\r);
 \fill (0,0) circle (0.05);

\draw (-2.2,0) -- (2.2,0);
\draw (0,-2.2) -- (0, 2.2);

\hunc{2}{
\uncover<2->{\draw[dblue, thick] (0,0) -- (2, 0) node[above right]{0$\degree$ = 0 rad};}
\uncover<3->{\draw[dblue, thick] (0,0) -- (-2,0) node[above left]{180$\degree$ \only<4->{= $\pi$ rad}};}
\uncover<5->{\draw[dblue, thick] (0,0) -- (0,2) node[above right] {90$\degree$ \only<6->{= $\frac{\pi}{2}$ rad}};}
\uncover<7->{\draw[dblue, thick] (0,0) -- (0,-2)  node[below right]{270$\degree$ \only<8->{= $\frac{3\pi}{2}$ rad}};}
\uncover<9->{\draw[dblue, thick] (45:0) -- (45:2) node[right] {45$\degree$ \only<10->{= $\frac{\pi}{4}$ rad}};}
\uncover<11->{\draw[dblue, thick] (30:0) -- (30:2) node[right] {30$\degree$ \only<12->{= $\frac{\pi}{6}$ rad}};}
\uncover<13->{\draw[dblue, thick] (60:0) -- (60:2) node[right] {60$\degree$ \only<14->{= $\frac{\pi}{3}$ rad}};}
 }
 \end{tikzpicture}
 \end{center}
 
 \end{frame}
 
 \begin{frame}
 
 \underline{Formule de la longueur d'arc}\\[0.2cm]
 
  \begin{minipage}{0.35\textwidth}
 
 \begin{tikzpicture}[scale=1]
 \draw[thick] (0,0) -- (-15:3) arc (-15:15:3) -- (0,0);
 \draw[dblue] (-15:3) arc (-15:15:3) node[right, yshift=-0.8cm] {$L_{\alpha}$};
 \draw[thick, ->, dred] (-15:0.5) arc (-15:15:0.5) node[below right, yshift=0.1cm]{$\alpha$};
 \draw[thick, dgreen] (0,0) -- node[above, sloped]{$r$} (15:3);
\end{tikzpicture}
 \end{minipage}\begin{minipage}{0.69\textwidth} 
 La \bl{longeur d'arc $L_{\alpha}$}interceptée par un \ve{angle de $\alpha$} sur un cercle de \ro{rayon $r$} est donnée par
  $$\boxed{\bl{L_{\alpha}} = \ro{r} \cdot \ve{\alpha}}$$ 
  \end{minipage}
  
\vspace{0.2cm}  
\uncover<2->{\ro{Cette formules n'est valable qui si l'angle est en radians!}}\\

 
\vfill 
  
\uncover<3->{  
\underline{Exemple 1.4} Trouver la longueur $L_{\alpha}$ sur la figure ci-dessous. \\[0.2cm]

 \begin{minipage}{0.35\textwidth}
 
 \begin{tikzpicture}[scale=1]
 \draw[thick] (0,0) -- (-15:3) arc (-15:15:3) -- (0,0);
 \draw[dblue] (-15:3) arc (-15:15:3) node[right, yshift=-0.8cm] {$L_{\alpha}$};
 \draw[thick, ->, dred] (-15:0.5) arc (-15:15:0.5) node[below right, yshift=0.1cm]{30\degree};
 \draw[thick, dgreen] (0,0) -- node[above, sloped]{40cm} (15:3);
\end{tikzpicture}
 \end{minipage}\begin{minipage}{0.69\textwidth}
\hunc{4}{Angle en radians: $\alpha = 30\degree \uncover<5->{= \dfrac{30\pi}{180}}\uncover<6->{ =  \dfrac{\pi}{6}}$. 

\uncover<7->{
On a:
\begin{eqnarray*}
\mathcal{L_{\alpha}} 
\uncover<8->{&=& r \cdot \alpha}\\
\uncover<9->{&=& 40 \cdot \dfrac{\pi}{6}}
\uncover<10->{= \dfrac{20\pi}{3}}
\end{eqnarray*}}
}
 \end{minipage}


}
 
 \end{frame}
 
 \begin{frame}
 
 \underline{Formule de l'aire d'un secteur}\\[0.2cm]
   
 \begin{minipage}{0.25\textwidth}
 
 \begin{tikzpicture}[scale=0.6]
 \draw[thick, fill=blue!30] (0,0) -- (-30:3) arc (-30:30:3) -- (0,0);
 \draw[thick, ->, dgreen] (-30:0.5) arc (-30:30:0.5) node[below right, yshift=0.1cm]{$\alpha$};
 \draw[thick, dred] (0,0) -- node[above, sloped]{r} (30:3);
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.74\textwidth}
 L'aire du secteur définit par l'angle $\alpha$ est donnée par: 
 $$\boxed{\bl{S_{\alpha}} = \frac{1}{2}\ro{r}^2\ve{\alpha}}$$ 
 \end{minipage}
 
 \vspace{0.2cm}
 \uncover<2->{\ro{Cette formule n'est valable qui si l'angle est en radians!}}
 
 \vfill
 \uncover<3->{
 \underline{Exemple 1.5} Trouver l'aire de la partie grisée sur la figure ci-dessous. \\
 
 \begin{minipage}{0.25\textwidth}
 
 \begin{tikzpicture}[scale=0.6]
 \draw[thick, fill=gray!50] (0,0) -- (-30:3) arc (-30:30:3) -- (0,0);
 \draw[thick, fill=white] (0,0) -- (-30:2) arc (-30:30:2) -- (0,0);
 \draw[thick, ->, dred] (-30:0.5) arc (-30:30:0.5) node[below right, yshift=0.1cm]{60\degree};
 \draw[thick, dgreen] (0,0) -- node[above, sloped]{30cm} (30:3);
 \draw[thick, dblue] (0,0) -- node[below, sloped]{20 cm}  (-30:2);
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.74\textwidth}
\hunc{4}{Angle en radians: $\alpha = 60\degree \uncover<5->{= \dfrac{60\pi}{180}}\uncover<6->{ =  \dfrac{\pi}{3}}$. 

\uncover<7->{
On a:
\begin{eqnarray*}
\mathcal{A} 
\uncover<8->{&=& \frac{1}{2}\cdot 30^2 \cdot \dfrac{\pi}{3}  -\frac{1}{2}\cdot 20^2 \cdot \dfrac{\pi}{3} }\\
\uncover<9->{&=& \dfrac{\pi}{6}(900-400) }
\uncover<10->{= \dfrac{500\pi}{6}}
\uncover<11->{= \dfrac{250\pi}{3} \mbox{ cm}^2}
\end{eqnarray*}
}}
 \end{minipage}}
 
 
 
 \end{frame}

 \def\ax{0}
 \def\ay{0}
 
 \def\bx{4}
 \def\by{0}
 
 \def\cx{4}
 \def\cy{3}
 
 
 \begin{frame}{2. Triangles rectangles}
 
 On définit les fonctions trigonométriques suivantes:\\[1cm]
 
 \begin{minipage}{0.48\textwidth}
 
 
 \bl{Sinus}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}[scale=0.5]
 
 \scriptsize
 
 \coordinate (A) at (\ax,\ay);
 \coordinate (B) at (\bx,\by);
 \coordinate (C) at (\cx,\cy);
 
 \draw (A) -- node[below] {adj} (B)  -- node[left] {opp} (C) -- node[above] {hyp}  (A);
 \node[above left] at (B) {$\square$};
 
 \draw (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);
 
 \hunc{2}{
 \draw[blue, thick] (B) -- (C); 
 \draw[blue, thick] (A) -- (C); 
 \draw[blue] (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);
 }
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.54\textwidth}
 
 \hunc{3}{$$\bl{\sin(\alpha) = \frac{\mbox{opp}}{\mbox{hyp}}}$$}
 
 \end{minipage}
 
 
 \end{minipage} \begin{minipage}{0.48\textwidth}
 
 \uncover<4->{
 
 \ro{Cosinus}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}[scale=0.5]
 
 \scriptsize
 
 \coordinate (A) at (\ax,\ay);
 \coordinate (B) at (\bx,\by);
 \coordinate (C) at (\cx,\cy);
 
 \draw (A) -- node[below] {adj} (B)  -- node[left] {opp} (C) -- node[above] {hyp}  (A);
 \node[above left] at (B) {$\square$};
 
 \draw (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);
 
 \hunc{5}{
 \draw[dred, thick] (A) -- (B); 
 \draw[dred, thick] (A) -- (C); 
 \draw[dred] (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);}
 
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.54\textwidth}
 
 \hunc{6}{$$\ro{\cos(\alpha) = \frac{\mbox{adj}}{\mbox{hyp}}}$$}
 
 \end{minipage}}
 
 
 \end{minipage}
 
 \vspace{1cm}
 
 \begin{center}
 \begin{minipage}{0.48\textwidth}
 
 \uncover<7->{ 

 \ve{Tangente}
 
 \begin{minipage}{0.44\textwidth}
 
 \begin{tikzpicture}[scale=0.5]
 
 \scriptsize
 
 \coordinate (A) at (\ax,\ay);
 \coordinate (B) at (\bx,\by);
 \coordinate (C) at (\cx,\cy);
 
 \draw (A) -- node[below] {adj} (B)  -- node[left] {opp} (C) -- node[above] {hyp}  (A);
 \node[above left] at (B) {$\square$};
 
 \draw (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);
 
 \hunc{8}{
 \draw[dgreen, thick] (B) -- (C); 
 \draw[dgreen, thick] (A) -- (B); 
 \draw[dgreen] (0:0.5) -- node[right, yshift=0.05cm] {$\alpha$} (36.8:0.5);}
 
 \end{tikzpicture}
 
 \end{minipage}\begin{minipage}{0.54\textwidth}
 
\hunc{9}{$$\ve{\tan(\alpha) = \frac{\mbox{opp}}{\mbox{adj}}}$$}
 
 \end{minipage}}
 
 \end{minipage} 
 \end{center}
  
 \end{frame}
 
 
 \begin{frame}
 
 \underline{Exemple 2.1} Résoudre le triangle rectangle dont on donne le côté $b = 8.2$ et l'angle $\alpha=37^\circ$. \\[0.5cm]
 
 \hunc{2}{
 \begin{minipage}{0.4\textwidth}
 
 \begin{center}
 \begin{tikzpicture}[scale=0.3]
 
 \coordinate (A) at (0,0);
 \coordinate (B) at (37:10.3);
 \coordinate (C) at (A -| B);
 
 \draw (A) node[below] {\bl{A}} -- node[below]{\ve{b=8.2}} (C) node[below] {\bl{C}} -- node[right]{\ve{$a$}}(B) node[above] {\bl{B}} 
	 -- node[above]{\ve{c}} (A);
 \draw (0:0.5) -- node[right, yshift=0.15cm] {$\ro{\alpha = 37^\circ}$} (37:0.5);
 \draw ($(B)+(-90:0.5)$) -- node[below, xshift=-0.15cm] {$\ro{\beta}$} ($(B)+(-143:0.5)$);
 
 \node[above left] at (C) {$\square$};
 
 \end{tikzpicture}
 \end{center}
 
 \end{minipage}\begin{minipage}{0.58\textwidth}
 
 \uncover<2->{\underline{Rappel} \bl{La somme des angles d'un triangle vaut 180$^\circ$ et son aire $$\boxed{\sigma=\dfrac{\mbox{ base } \cdot \mbox{ hauteur}}{2}}$$ }}\\
  
 \end{minipage}
 
 \vspace{0.5cm}
 
  \uncover<3->{
 On a
 $$\beta = \uncover<4->{180 - 90 - 37 = \uncover<5->{53^\circ}}$$}
 
 \uncover<6->{
 Pour les côtés:
 \begin{itemize}
 \item<7-> $\tan(\alpha) = \frac{a}{b} \uncover<8->{ \Rightarrow \tan(37^\circ) = \frac{a}{8.2}}$
 \uncover<9->{$\Rightarrow a = 8.2\cdot\tan(37^\circ) \uncover<10->{\approx 6.2}$}
 \item<11-> $\cos(\alpha) = \frac{b}{c} \uncover<12->{ \Rightarrow \cos(37^\circ) = \frac{8.2}{c}}$
 \uncover<13->{$\Rightarrow c = \frac{8.2}{\cos(37^\circ)} \uncover<14->{\approx 10.3}$}
 \end{itemize}}
 \uncover<15->{Pour l'aire: $\sigma = \frac{b \cdot a}{2} \uncover<16->{\approx \frac{8.2 \cdot 6.2}{2} = 25.3}$}}
 
 \end{frame}
 
 \begin{frame}
 
\underline{Exemple 2.2} Résoudre le triangle rectangle dont on donne le côté $a = 7.6$ et l'hypothénuse $c= 18.4$. \\[0.5cm]
 
 \hunc{2}{
 \begin{minipage}{0.4\textwidth}
 
 \begin{center}
 \begin{tikzpicture}[scale=0.3]
 
 \coordinate (A) at (0,0);
 \coordinate (B) at (37:10.3);
 \coordinate (C) at (A -| B);
 
 \draw (A) node[below] {\bl{A}} -- node[below]{\ve{b}} (C) node[below] {\bl{C}} -- node[above, sloped]{\ve{$a=7.6$}}(B) node[above] {\bl{B}} 
	 -- node[above,sloped]{\ve{c=18.4}} (A);
 \draw (0:0.5) -- node[right, yshift=0.15cm] {$\ro{\alpha}$} (37:0.5);
 \draw ($(B)+(-90:0.5)$) -- node[below, xshift=-0.15cm] {$\ro{\beta}$} ($(B)+(-143:0.5)$);
 
 \node[above left] at (C) {$\square$};
 
 \end{tikzpicture}
 \end{center}
 
 \end{minipage}\begin{minipage}{0.58\textwidth}
 
 \uncover<2->{\underline{Rappel} \bl{Théorème de Pythagore $$\boxed{a^2 + b^2 = c^2}$$ }}\\
  
 \end{minipage}
 
 \vspace{0.5cm}
 
  \uncover<3->{
 On a
 $$b \uncover<4->{= \sqrt{c^2 - a^2}} \uncover<5->{= \sqrt{18.4^2 - 7.6^2}} \uncover<6->{\approx 16.8}$$}
 
 \uncover<7->{
 Pour les angles:
 \begin{itemize}
 \item<8-> $\sin(\alpha) = \frac{a}{c} \uncover<9->{= \frac{7.6}{18.4}$}
 \uncover<10->{$\Rightarrow \alpha = \sin^{-1}\left(\frac{7.6}{18.4}\right) \uncover<11->{\approx 24.4^\circ}$}
 \item<12-> $\beta \uncover<13->{= 180 - 90 - 24.4 \uncover<14->{\approx 65.6^\circ}}$. 
 \end{itemize}}
 
  \uncover<15->{Pour l'aire: $\sigma = \frac{b \cdot a}{2} \uncover<16->{= \frac{16.8 \cdot 7.6}{2} \approx 6.7}$}}
 
 \end{frame}
 
 
\end{document}




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