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\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle AMB rectangle en A, on sait que :
\begin{itemize}
\item MB=0,8 cm
\item \(\widehat{MBA}\) =  32\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [AM]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle SBJ rectangle en S, on sait que :
\begin{itemize}
\item SJ=1,3 cm
\item \(\widehat{SBJ}\) = 78\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [JB]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle BSW rectangle en B, on sait que :
\begin{itemize}
\item BW=6,2 cm
\item SW=7,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BWS}\).

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle GAH rectangle en G, on sait que :
\begin{itemize}
\item GA=2,6 cm
\item \(\widehat{GAH}\) = 67\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GH]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle TSL rectangle en T, on sait que :
\begin{itemize}
\item TS=2,9 cm
\item TL=4,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TSL}\). \newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

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 Dans le triangle AMB rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{ABM}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{ABM}\)) = \(\dfrac{AM}{MB}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{AM}{0{,}8}\)

On a donc AM = 0,8 \(\times\) sin(32\textdegree{}) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle SBJ rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SBJ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{SBJ}\)) =  \(\dfrac{SJ}{BJ}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{1{,}3}{BJ}\)
\par\addvspace{.5em}
 On a donc BJ = \(\dfrac{1{,}3}{\sin(78^{\circ})}\) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


 Dans le triangle BSW rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BWS}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{BWS}\)) = \(\dfrac{BW}{SW}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BWS}\)) =  \(\dfrac{6{,}2}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{BWS}\) = Arccos (\(\dfrac{6{,}2}{7{,}7}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle GAH rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GAH}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{GAH}\)) = \(\dfrac{GH}{GA}\)
\par\addvspace{.5em}
d'où tan(67\textdegree{}) =\(\dfrac{GH}{2{,}6}\)
\par\addvspace{.5em}
On a donc GH = 2,6 \(\times\) tan(67\textdegree{}) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.782,2.743) {?};
\end{tikzpicture}\end{center}


 Dans le triangle TSL rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TSL}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TSL}\)) = \(\dfrac{TL}{TS}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{TSL}\)) = \(\dfrac{4{,}2}{2{,}9}\)
\par\addvspace{.5em}
 On a donc TSL = ArcTan(\(\dfrac{4{,}2}{2{,}9}\)) \(\approx\) 55\textdegree{}\end{document}
