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\begin{document}

\sloppy

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

 Dans le triangle ARS rectangle en A, on sait que :
\begin{itemize}
\item AS=6,7 cm
\item \(\widehat{ARS}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle HML rectangle en H, on sait que :
\begin{itemize}
\item HM=2,2 cm
\item ML=7,5 cm
\end{itemize}

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

 Dans le triangle HPR rectangle en H, on sait que :
\begin{itemize}
\item HP=2,2 cm
\item HR=4,2 cm
\end{itemize}

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

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

 Dans le triangle DJN rectangle en D, on sait que :
\begin{itemize}
\item JN=9,5 cm
\item \(\widehat{JND}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle MNK rectangle en M, on sait que :
\begin{itemize}
\item MK=6,9 cm
\item \(\widehat{NKM}\) =  44\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ARS}\)) = \(\dfrac{AS}{AR}\)
\par\addvspace{.5em}
tan(57\textdegree{}) = \(\dfrac{6{,}7}{AR}\)
\par\addvspace{.5em}
 On a donc AR = \(\dfrac{6{,}7}{\tan(57^{\circ})}\) \(\approx\) 4,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{HML}\)) = \(\dfrac{HM}{ML}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HML}\)) = \(\dfrac{2{,}2}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{HML}\) =ArcCos(\(\dfrac{2{,}2}{7{,}5}\)) \(\approx\)73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HRP}\)) = \(\dfrac{HP}{HR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HRP}\)) =  \(\dfrac{2{,}2}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HRP}\) =  ArcTan(\(\dfrac{2{,}2}{4{,}2}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{DNJ}\)) = \(\dfrac{DN}{JN}\)
\par\addvspace{.5em}
  d'où    cos(17\textdegree{}) =  \(\dfrac{DN}{9{,}5}\)

On a donc DN = 9,5 \(\times\) cos(17\textdegree{}) \(\approx\) 9{,}1 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MKN}\)) = \(\dfrac{MK}{NK}\)
\par\addvspace{.5em}
  d'où    cos(44\textdegree{}) =  \(\dfrac{6{,}9}{NK}\)
\par\addvspace{.5em}
 On a donc NK =  \(\dfrac{6{,}9}{\cos(44^{\circ})}\) \(\approx\) 9,6 cm\end{document}
