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\begin{document}

\sloppy

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

 Dans le triangle ANZ rectangle en A, on sait que :
\begin{itemize}
\item NZ=8,3 cm
\item \(\widehat{NZA}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle LNJ rectangle en L, on sait que :
\begin{itemize}
\item LN=1,4 cm
\item LJ=4,8 cm
\end{itemize}

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

 Dans le triangle NRA rectangle en N, on sait que :
\begin{itemize}
\item RA=2,1 cm
\item \(\widehat{RAN}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle SLM rectangle en S, on sait que :
\begin{itemize}
\item SL=1,7 cm
\item SM=5,3 cm
\end{itemize}

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

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

 Dans le triangle AHM rectangle en A, on sait que :
\begin{itemize}
\item AH=2,2 cm
\item \(\widehat{HMA}\) =  24\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{AZN}\)) = \(\dfrac{AN}{NZ}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{AN}{8{,}3}\)

On a donc AN = 8,3 \(\times\) sin(31\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LNJ}\)) = \(\dfrac{LJ}{LN}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{LNJ}\)) = \(\dfrac{4{,}8}{1{,}4}\)
\par\addvspace{.5em}
 On a donc LNJ = ArcTan(\(\dfrac{4{,}8}{1{,}4}\)) \(\approx\) 74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{NAR}\)) = \(\dfrac{NA}{RA}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{NA}{2{,}1}\)

On a donc NA = 2,1 \(\times\) cos(38\textdegree{}) \(\approx\) 1{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{SML}\)) = \(\dfrac{SL}{SM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SML}\)) =  \(\dfrac{1{,}7}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SML}\) =  ArcTan(\(\dfrac{1{,}7}{5{,}3}\)) \(\approx\) 18\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.298,0.981) {24°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AMH}\)) = \(\dfrac{AH}{HM}\)
\par\addvspace{.5em}
 D'où sin(24\textdegree{}) = \(\dfrac{2{,}2}{HM}\)
\par\addvspace{.5em}
On a donc HM =\(\dfrac{2{,}2}{\sin(24^{\circ})}\) \(\approx\) 5,4 cm\end{document}
