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\begin{document}

\sloppy

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

 Dans le triangle CNG rectangle en C, on sait que :
\begin{itemize}
\item CG=1,8 cm
\item \(\widehat{CNG}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle SVH rectangle en S, on sait que :
\begin{itemize}
\item SV=4,5 cm
\item \(\widehat{VHS}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle CJR rectangle en C, on sait que :
\begin{itemize}
\item CR=8,5 cm
\item \(\widehat{CJR}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle NDZ rectangle en N, on sait que :
\begin{itemize}
\item ND=2,4 cm
\item NZ=4,4 cm
\end{itemize}

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

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

 Dans le triangle HFN rectangle en H, on sait que :
\begin{itemize}
\item HF=3,2 cm
\item FN=7,7 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CNG}\)) =  \(\dfrac{CG}{NG}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{1{,}8}{NG}\)
\par\addvspace{.5em}
 On a donc NG = \(\dfrac{1{,}8}{\sin(68^{\circ})}\) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{SHV}\)) = \(\dfrac{SV}{SH}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{4{,}5}{SH}\)
\par\addvspace{.5em}
On a donc SH = \(\dfrac{4{,}5}{\tan(38^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{CJR}\)) = \(\dfrac{CR}{CJ}\)
\par\addvspace{.5em}
tan(53\textdegree{}) = \(\dfrac{8{,}5}{CJ}\)
\par\addvspace{.5em}
 On a donc CJ = \(\dfrac{8{,}5}{\tan(53^{\circ})}\) \(\approx\) 6,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NZD}\)) = \(\dfrac{ND}{NZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NZD}\)) =  \(\dfrac{2{,}4}{4{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NZD}\) =  ArcTan(\(\dfrac{2{,}4}{4{,}4}\)) \(\approx\) 29\textdegree{}
\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) {7,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.410,2.492) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HFN}\)) = \(\dfrac{HF}{FN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HFN}\)) = \(\dfrac{3{,}2}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{HFN}\) =ArcCos(\(\dfrac{3{,}2}{7{,}7}\)) \(\approx\)65\textdegree{}\end{document}
