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\begin{document}

\sloppy

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

 Dans le triangle HRC rectangle en H, on sait que :
\begin{itemize}
\item HR=6,8 cm
\item \(\widehat{HRC}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle JAS rectangle en J, on sait que :
\begin{itemize}
\item JA=2,6 cm
\item AS=9,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{JAS}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle GMN rectangle en G, on sait que :
\begin{itemize}
\item MN=5,6 cm
\item \(\widehat{GMN}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle CMG rectangle en C, on sait que :
\begin{itemize}
\item CG=5 cm
\item \(\widehat{CMG}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZCN rectangle en Z, on sait que :
\begin{itemize}
\item ZC=2,9 cm
\item CN=9,7 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HRC}\)) = \(\dfrac{HC}{HR}\)
\par\addvspace{.5em}
d'où tan(50\textdegree{}) =\(\dfrac{HC}{6{,}8}\)
\par\addvspace{.5em}
On a donc HC = 6,8 \(\times\) tan(50\textdegree{}) \(\approx\) 8,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{JAS}\)) = \(\dfrac{JA}{AS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JAS}\)) = \(\dfrac{2{,}6}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{JAS}\) =ArcCos(\(\dfrac{2{,}6}{9{,}9}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{GMN}\)) = \(\dfrac{GM}{MN}\)
\par\addvspace{.5em}
d'où cos(71\textdegree{}) = \(\dfrac{GM}{5{,}6}\)

On a donc GM = 5,6 \(\times\) cos(71\textdegree{}) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CMG}\)) =  \(\dfrac{CG}{MG}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{5}{MG}\)
\par\addvspace{.5em}
 On a donc MG = \(\dfrac{5}{\sin(71^{\circ})}\) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.267) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZNC}\)) = \(\dfrac{ZC}{CN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZNC}\)) =\(\dfrac{2{,}9}{9{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZNC}\) = ArcSin(\(\dfrac{2{,}9}{9{,}7}\)) \(\approx\) 17\textdegree{}\end{document}
