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\begin{document}

\sloppy

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

 Dans le triangle MAJ rectangle en M, on sait que :
\begin{itemize}
\item MJ=4,7 cm
\item \(\widehat{MAJ}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle NML rectangle en N, on sait que :
\begin{itemize}
\item NM=4,6 cm
\item \(\widehat{NML}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle NVT rectangle en N, on sait que :
\begin{itemize}
\item NT=4,7 cm
\item VT=9,3 cm
\end{itemize}

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

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

 Dans le triangle AJP rectangle en A, on sait que :
\begin{itemize}
\item AJ=1 cm
\item AP=5,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AJP}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle BJW rectangle en B, on sait que :
\begin{itemize}
\item BW=2,4 cm
\item \(\widehat{BJW}\) = 62\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{MAJ}\)) =  \(\dfrac{MJ}{AJ}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{4{,}7}{AJ}\)
\par\addvspace{.5em}
 On a donc AJ = \(\dfrac{4{,}7}{\sin(73^{\circ})}\) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NML}\)) = \(\dfrac{NL}{NM}\)
\par\addvspace{.5em}
d'où tan(80\textdegree{}) =\(\dfrac{NL}{4{,}6}\)
\par\addvspace{.5em}
On a donc NL = 4,6 \(\times\) tan(80\textdegree{}) \(\approx\) 26,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{NTV}\)) = \(\dfrac{NT}{VT}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{NTV}\)) =  \(\dfrac{4{,}7}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{NTV}\) = Arccos (\(\dfrac{4{,}7}{9{,}3}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{AJP}\)) = \(\dfrac{AP}{AJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{AJP}\)) = \(\dfrac{5{,}7}{1}\)
\par\addvspace{.5em}
 On a donc AJP = ArcTan(5,7) \(\approx\) 80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BJW}\)) = \(\dfrac{BW}{BJ}\)
\par\addvspace{.5em}
tan(62\textdegree{}) = \(\dfrac{2{,}4}{BJ}\)
\par\addvspace{.5em}
 On a donc BJ = \(\dfrac{2{,}4}{\tan(62^{\circ})}\) \(\approx\) 1,3 cm\end{document}
