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\begin{document}

\sloppy

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

 Dans le triangle CKJ rectangle en C, on sait que :
\begin{itemize}
\item KJ=8,1 cm
\item \(\widehat{CKJ}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSA rectangle en K, on sait que :
\begin{itemize}
\item KA=5,6 cm
\item SA=9,3 cm
\end{itemize}

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

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

 Dans le triangle JLF rectangle en J, on sait que :
\begin{itemize}
\item JL=1,7 cm
\item LF=7,6 cm
\end{itemize}

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

 Dans le triangle RTH rectangle en R, on sait que :
\begin{itemize}
\item TH=4,4 cm
\item \(\widehat{RTH}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle MSC rectangle en M, on sait que :
\begin{itemize}
\item MC=7,1 cm
\item \(\widehat{SCM}\) =  44\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CKJ}\)) =  \(\dfrac{CJ}{KJ}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{CJ}{8{,}1}\)

On a donc CJ = 8,1 \(\times\) sin(51\textdegree{}) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KAS}\)) = \(\dfrac{KA}{SA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KAS}\)) =  \(\dfrac{5{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{KAS}\) = Arccos (\(\dfrac{5{,}6}{9{,}3}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{JLF}\)) = \(\dfrac{JL}{LF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JLF}\)) = \(\dfrac{1{,}7}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{JLF}\) =ArcCos(\(\dfrac{1{,}7}{7{,}6}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{RTH}\)) = \(\dfrac{RT}{TH}\)
\par\addvspace{.5em}
d'où cos(64\textdegree{}) = \(\dfrac{RT}{4{,}4}\)

On a donc RT = 4,4 \(\times\) cos(64\textdegree{}) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{MCS}\)) = \(\dfrac{MC}{SC}\)
\par\addvspace{.5em}
  d'où    cos(44\textdegree{}) =  \(\dfrac{7{,}1}{SC}\)
\par\addvspace{.5em}
 On a donc SC =  \(\dfrac{7{,}1}{\cos(44^{\circ})}\) \(\approx\) 9,9 cm\end{document}
