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\begin{document}

\sloppy

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

 Dans le triangle SJA rectangle en S, on sait que :
\begin{itemize}
\item JA=3,5 cm
\item \(\widehat{JAS}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle PNR rectangle en P, on sait que :
\begin{itemize}
\item PR=6,5 cm
\item NR=9,7 cm
\end{itemize}

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

 Dans le triangle VWN rectangle en V, on sait que :
\begin{itemize}
\item VW=5,4 cm
\item \(\widehat{WNV}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle GZV rectangle en G, on sait que :
\begin{itemize}
\item ZV=4,4 cm
\item \(\widehat{GZV}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle PBR rectangle en P, on sait que :
\begin{itemize}
\item PB=1,2 cm
\item BR=7,3 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SAJ}\)) = \(\dfrac{SJ}{JA}\)
\par\addvspace{.5em}
 D'où sin(11\textdegree{}) = \(\dfrac{SJ}{3{,}5}\)

On a donc SJ = 3,5 \(\times\) sin(11\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{PNR}\)) =  \(\dfrac{PR}{NR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PNR}\)) = \(\dfrac{6{,}5}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{PNR}\) =  ArcSin(\(\dfrac{6{,}5}{9{,}7}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VNW}\)) = \(\dfrac{VW}{WN}\)
\par\addvspace{.5em}
 D'où sin(27\textdegree{}) = \(\dfrac{5{,}4}{WN}\)
\par\addvspace{.5em}
On a donc WN =\(\dfrac{5{,}4}{\sin(27^{\circ})}\) \(\approx\) 11,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GZV}\)) =  \(\dfrac{GV}{ZV}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{GV}{4{,}4}\)

On a donc GV = 4,4 \(\times\) sin(71\textdegree{}) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.580) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PRB}\)) = \(\dfrac{PB}{BR}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PRB}\)) =\(\dfrac{1{,}2}{7{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PRB}\) = ArcSin(\(\dfrac{1{,}2}{7{,}3}\)) \(\approx\) 9\textdegree{}\end{document}
