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\begin{document}

\sloppy

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

 Dans le triangle ZML rectangle en Z, on sait que :
\begin{itemize}
\item ZM=8,3 cm
\item \(\widehat{ZML}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle RDK rectangle en R, on sait que :
\begin{itemize}
\item RD=1,9 cm
\item RK=6,4 cm
\end{itemize}

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

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

 Dans le triangle NHA rectangle en N, on sait que :
\begin{itemize}
\item NA=4,6 cm
\item \(\widehat{NHA}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle CTL rectangle en C, on sait que :
\begin{itemize}
\item CL=2,5 cm
\item \(\widehat{TLC}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle ARW rectangle en A, on sait que :
\begin{itemize}
\item AW=5,1 cm
\item RW=7,6 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ZML}\)) = \(\dfrac{ZL}{ZM}\)
\par\addvspace{.5em}
d'où tan(58\textdegree{}) =\(\dfrac{ZL}{8{,}3}\)
\par\addvspace{.5em}
On a donc ZL = 8,3 \(\times\) tan(58\textdegree{}) \(\approx\) 13,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{RKD}\)) = \(\dfrac{RD}{RK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RKD}\)) =  \(\dfrac{1{,}9}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RKD}\) =  ArcTan(\(\dfrac{1{,}9}{6{,}4}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{NHA}\)) =  \(\dfrac{NA}{HA}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{4{,}6}{HA}\)
\par\addvspace{.5em}
 On a donc HA = \(\dfrac{4{,}6}{\sin(50^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CLT}\)) = \(\dfrac{CT}{CL}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{CT}{2{,}5}\)
\par\addvspace{.5em}
On a donc CT =  2,5 \(\times\) tan(10\textdegree{})  \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.473,2.711) {?};
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 Dans le triangle ARW rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{ARW}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{ARW}\)) =  \(\dfrac{AW}{RW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ARW}\)) = \(\dfrac{5{,}1}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{ARW}\) =  ArcSin(\(\dfrac{5{,}1}{7{,}6}\)) \(\approx\) 42\textdegree{}\end{document}
