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\begin{document}

\sloppy

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

 Dans le triangle MCS rectangle en M, on sait que :
\begin{itemize}
\item MC=1,5 cm
\item CS=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MCS}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle PJZ rectangle en P, on sait que :
\begin{itemize}
\item PZ=8,7 cm
\item \(\widehat{JZP}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle FRJ rectangle en F, on sait que :
\begin{itemize}
\item FR=2,7 cm
\item RJ=7 cm
\end{itemize}

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

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

 Dans le triangle FSL rectangle en F, on sait que :
\begin{itemize}
\item FS=4,3 cm
\item \(\widehat{SLF}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle DJP rectangle en D, on sait que :
\begin{itemize}
\item DP=5 cm
\item \(\widehat{JPD}\) =  41\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{MCS}\)) = \(\dfrac{MC}{CS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MCS}\)) = \(\dfrac{1{,}5}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{MCS}\) =ArcCos(\(\dfrac{1{,}5}{9{,}4}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PZJ}\)) = \(\dfrac{PJ}{PZ}\)
\par\addvspace{.5em}
d'où  tan(34\textdegree{}) =  \(\dfrac{PJ}{8{,}7}\)
\par\addvspace{.5em}
On a donc PJ =  8,7 \(\times\) tan(34\textdegree{})  \(\approx\) 5,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FJR}\)) = \(\dfrac{FR}{RJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FJR}\)) =\(\dfrac{2{,}7}{7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FJR}\) = ArcSin(\(\dfrac{2{,}7}{7}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FLS}\)) = \(\dfrac{FS}{FL}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{4{,}3}{FL}\)
\par\addvspace{.5em}
On a donc FL = \(\dfrac{4{,}3}{\tan(19^{\circ})}\) \(\approx\) 12,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5 cm};
  \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 (4.486,0.767) {41°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DPJ}\)) = \(\dfrac{DP}{JP}\)
\par\addvspace{.5em}
  d'où    cos(41\textdegree{}) =  \(\dfrac{5}{JP}\)
\par\addvspace{.5em}
 On a donc JP =  \(\dfrac{5}{\cos(41^{\circ})}\) \(\approx\) 6,6 cm\end{document}
