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\begin{document}

\sloppy

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

 Dans le triangle GKS rectangle en G, on sait que :
\begin{itemize}
\item GK=2,3 cm
\item GS=6,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GKS}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle TBG rectangle en T, on sait que :
\begin{itemize}
\item TG=8,2 cm
\item \(\widehat{BGT}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJD rectangle en M, on sait que :
\begin{itemize}
\item JD=4,7 cm
\item \(\widehat{MJD}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle CSG rectangle en C, on sait que :
\begin{itemize}
\item SG=5,3 cm
\item \(\widehat{CSG}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZJC rectangle en Z, on sait que :
\begin{itemize}
\item ZJ=1,1 cm
\item ZC=5,9 cm
\end{itemize}

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

\newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{GKS}\)) = \(\dfrac{GS}{GK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{GKS}\)) = \(\dfrac{6{,}4}{2{,}3}\)
\par\addvspace{.5em}
 On a donc GKS = ArcTan(\(\dfrac{6{,}4}{2{,}3}\)) \(\approx\) 70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{TGB}\)) = \(\dfrac{TG}{BG}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{8{,}2}{BG}\)
\par\addvspace{.5em}
 On a donc BG =  \(\dfrac{8{,}2}{\cos(38^{\circ})}\) \(\approx\) 10,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MJD}\)) = \(\dfrac{MJ}{JD}\)
\par\addvspace{.5em}
d'où cos(76\textdegree{}) = \(\dfrac{MJ}{4{,}7}\)

On a donc MJ = 4,7 \(\times\) cos(76\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CSG}\)) =  \(\dfrac{CG}{SG}\)
\par\addvspace{.5em}
 d'où sin(59\textdegree{}) = \(\dfrac{CG}{5{,}3}\)

On a donc CG = 5,3 \(\times\) sin(59\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.210) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.538) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZCJ}\)) = \(\dfrac{ZJ}{ZC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZCJ}\)) =  \(\dfrac{1{,}1}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZCJ}\) =  ArcTan(\(\dfrac{1{,}1}{5{,}9}\)) \(\approx\) 11\textdegree{}\end{document}
