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\begin{document}

\sloppy

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

 Dans le triangle NCZ rectangle en N, on sait que :
\begin{itemize}
\item NZ=3,2 cm
\item \(\widehat{NCZ}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle HJV rectangle en H, on sait que :
\begin{itemize}
\item HV=4,8 cm
\item \(\widehat{HJV}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle NLD rectangle en N, on sait que :
\begin{itemize}
\item NL=2 cm
\item ND=6,5 cm
\end{itemize}

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

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

 Dans le triangle SHC rectangle en S, on sait que :
\begin{itemize}
\item HC=6,8 cm
\item \(\widehat{HCS}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle VNG rectangle en V, on sait que :
\begin{itemize}
\item VN=2,9 cm
\item NG=6,7 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NCZ}\)) = \(\dfrac{NZ}{NC}\)
\par\addvspace{.5em}
tan(60\textdegree{}) = \(\dfrac{3{,}2}{NC}\)
\par\addvspace{.5em}
 On a donc NC = \(\dfrac{3{,}2}{\tan(60^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HJV}\)) =  \(\dfrac{HV}{JV}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{4{,}8}{JV}\)
\par\addvspace{.5em}
 On a donc JV = \(\dfrac{4{,}8}{\sin(70^{\circ})}\) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NDL}\)) = \(\dfrac{NL}{ND}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NDL}\)) =  \(\dfrac{2}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NDL}\) =  ArcTan(\(\dfrac{2}{6{,}5}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SCH}\)) = \(\dfrac{SC}{HC}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{SC}{6{,}8}\)

On a donc SC = 6,8 \(\times\) cos(37\textdegree{}) \(\approx\) 5{,}4 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) {6,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.405,2.545) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VNG}\)) = \(\dfrac{VN}{NG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VNG}\)) = \(\dfrac{2{,}9}{6{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{VNG}\) =ArcCos(\(\dfrac{2{,}9}{6{,}7}\)) \(\approx\)64\textdegree{}\end{document}
