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\begin{document}

\sloppy

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

 Dans le triangle NMR rectangle en N, on sait que :
\begin{itemize}
\item MR=6,9 cm
\item \(\widehat{NMR}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle SFM rectangle en S, on sait que :
\begin{itemize}
\item SF=3 cm
\item \(\widehat{FMS}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle LNT rectangle en L, on sait que :
\begin{itemize}
\item LN=2,4 cm
\item NT=9,2 cm
\end{itemize}

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

 Dans le triangle CJA rectangle en C, on sait que :
\begin{itemize}
\item CJ=7,8 cm
\item \(\widehat{JAC}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle NZP rectangle en N, on sait que :
\begin{itemize}
\item NZ=3,3 cm
\item NP=4,8 cm
\end{itemize}

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

\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 (1.808,2.742) {55°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{NMR}\)) = \(\dfrac{NM}{MR}\)
\par\addvspace{.5em}
d'où cos(55\textdegree{}) = \(\dfrac{NM}{6{,}9}\)

On a donc NM = 6,9 \(\times\) cos(55\textdegree{}) \(\approx\) 4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{SMF}\)) = \(\dfrac{SF}{FM}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{3}{FM}\)
\par\addvspace{.5em}
On a donc FM =\(\dfrac{3}{\sin(32^{\circ})}\) \(\approx\) 5,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{LNT}\)) = \(\dfrac{LN}{NT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LNT}\)) = \(\dfrac{2{,}4}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{LNT}\) =ArcCos(\(\dfrac{2{,}4}{9{,}2}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CAJ}\)) = \(\dfrac{CJ}{CA}\)
\par\addvspace{.5em}
d'où  tan(24\textdegree{}) =  \(\dfrac{7{,}8}{CA}\)
\par\addvspace{.5em}
On a donc CA = \(\dfrac{7{,}8}{\tan(24^{\circ})}\) \(\approx\) 17,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.900,0.731) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NPZ}\)) = \(\dfrac{NZ}{NP}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NPZ}\)) =  \(\dfrac{3{,}3}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NPZ}\) =  ArcTan(\(\dfrac{3{,}3}{4{,}8}\)) \(\approx\) 35\textdegree{}\end{document}
