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\begin{document}

\sloppy

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

 Dans le triangle MNZ rectangle en M, on sait que :
\begin{itemize}
\item MN=8,7 cm
\item \(\widehat{NZM}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle JDF rectangle en J, on sait que :
\begin{itemize}
\item JF=5,7 cm
\item DF=7 cm
\end{itemize}

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

 Dans le triangle RPS rectangle en R, on sait que :
\begin{itemize}
\item RP=2,2 cm
\item PS=8,7 cm
\end{itemize}

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

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

 Dans le triangle AFC rectangle en A, on sait que :
\begin{itemize}
\item AC=9 cm
\item \(\widehat{AFC}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSJ rectangle en K, on sait que :
\begin{itemize}
\item KS=1,5 cm
\item \(\widehat{SJK}\) =  42\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MZN}\)) = \(\dfrac{MN}{MZ}\)
\par\addvspace{.5em}
d'où  tan(44\textdegree{}) =  \(\dfrac{8{,}7}{MZ}\)
\par\addvspace{.5em}
On a donc MZ = \(\dfrac{8{,}7}{\tan(44^{\circ})}\) \(\approx\) 9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{JDF}\)) =  \(\dfrac{JF}{DF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{JDF}\)) = \(\dfrac{5{,}7}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{JDF}\) =  ArcSin(\(\dfrac{5{,}7}{7}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{RSP}\)) = \(\dfrac{RP}{PS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RSP}\)) =\(\dfrac{2{,}2}{8{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RSP}\) = ArcSin(\(\dfrac{2{,}2}{8{,}7}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{AFC}\)) = \(\dfrac{AC}{AF}\)
\par\addvspace{.5em}
tan(74\textdegree{}) = \(\dfrac{9}{AF}\)
\par\addvspace{.5em}
 On a donc AF = \(\dfrac{9}{\tan(74^{\circ})}\) \(\approx\) 2,6 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 (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KJS}\)) = \(\dfrac{KS}{SJ}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{1{,}5}{SJ}\)
\par\addvspace{.5em}
On a donc SJ =\(\dfrac{1{,}5}{\sin(42^{\circ})}\) \(\approx\) 2,2 cm\end{document}
