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\begin{document}

\sloppy

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

 Dans le triangle RDM rectangle en R, on sait que :
\begin{itemize}
\item RM=8,6 cm
\item \(\widehat{RDM}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle SCL rectangle en S, on sait que :
\begin{itemize}
\item SL=0,6 cm
\item \(\widehat{CLS}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRJ rectangle en T, on sait que :
\begin{itemize}
\item TJ=4,9 cm
\item RJ=7,5 cm
\end{itemize}

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

 Dans le triangle DZC rectangle en D, on sait que :
\begin{itemize}
\item DZ=2,2 cm
\item DC=5,9 cm
\end{itemize}

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

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

 Dans le triangle JHZ rectangle en J, on sait que :
\begin{itemize}
\item HZ=1,2 cm
\item \(\widehat{HZJ}\) =  41\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RDM}\)) = \(\dfrac{RM}{RD}\)
\par\addvspace{.5em}
tan(55\textdegree{}) = \(\dfrac{8{,}6}{RD}\)
\par\addvspace{.5em}
 On a donc RD = \(\dfrac{8{,}6}{\tan(55^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SLC}\)) = \(\dfrac{SL}{CL}\)
\par\addvspace{.5em}
  d'où    cos(30\textdegree{}) =  \(\dfrac{0{,}6}{CL}\)
\par\addvspace{.5em}
 On a donc CL =  \(\dfrac{0{,}6}{\cos(30^{\circ})}\) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle TRJ rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TRJ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TRJ}\)) =  \(\dfrac{TJ}{RJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{TRJ}\)) = \(\dfrac{4{,}9}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{TRJ}\) =  ArcSin(\(\dfrac{4{,}9}{7{,}5}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DCZ}\)) = \(\dfrac{DZ}{DC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DCZ}\)) =  \(\dfrac{2{,}2}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DCZ}\) =  ArcTan(\(\dfrac{2{,}2}{5{,}9}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.486,0.767) {41°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JZH}\)) = \(\dfrac{JZ}{HZ}\)
\par\addvspace{.5em}
  d'où    cos(41\textdegree{}) =  \(\dfrac{JZ}{1{,}2}\)

On a donc JZ = 1,2 \(\times\) cos(41\textdegree{}) \(\approx\) 0{,}9 cm\end{document}
