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\begin{document}

\sloppy

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

 Dans le triangle JHG rectangle en J, on sait que :
\begin{itemize}
\item JH=7,1 cm
\item \(\widehat{JHG}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle NRF rectangle en N, on sait que :
\begin{itemize}
\item NF=5,2 cm
\item RF=7,7 cm
\end{itemize}

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

 Dans le triangle GNJ rectangle en G, on sait que :
\begin{itemize}
\item GN=2,9 cm
\item NJ=8,9 cm
\end{itemize}

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

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

 Dans le triangle WLA rectangle en W, on sait que :
\begin{itemize}
\item LA=10 cm
\item \(\widehat{LAW}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle JMA rectangle en J, on sait que :
\begin{itemize}
\item JA=4,8 cm
\item \(\widehat{MAJ}\) =  25\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{JHG}\)) = \(\dfrac{JG}{JH}\)
\par\addvspace{.5em}
d'où tan(51\textdegree{}) =\(\dfrac{JG}{7{,}1}\)
\par\addvspace{.5em}
On a donc JG = 7,1 \(\times\) tan(51\textdegree{}) \(\approx\) 8,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NRF}\)) =  \(\dfrac{NF}{RF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NRF}\)) = \(\dfrac{5{,}2}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{NRF}\) =  ArcSin(\(\dfrac{5{,}2}{7{,}7}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{GJN}\)) = \(\dfrac{GN}{NJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GJN}\)) =\(\dfrac{2{,}9}{8{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GJN}\) = ArcSin(\(\dfrac{2{,}9}{8{,}9}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.302,0.841) {27°};
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 Dans le triangle WLA rectangle en W, on cherche une relation entre l'angle aigu   \(\widehat{WAL}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{WAL}\)) = \(\dfrac{WL}{LA}\)
\par\addvspace{.5em}
 D'où sin(27\textdegree{}) = \(\dfrac{WL}{10}\)

On a donc WL = 10 \(\times\) sin(27\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.299,0.935) {25°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JAM}\)) = \(\dfrac{JA}{MA}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{4{,}8}{MA}\)
\par\addvspace{.5em}
 On a donc MA =  \(\dfrac{4{,}8}{\cos(25^{\circ})}\) \(\approx\) 5,3 cm\end{document}
