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\begin{document}

\sloppy

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

 Dans le triangle GRA rectangle en G, on sait que :
\begin{itemize}
\item GA=4,9 cm
\item \(\widehat{GRA}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGF rectangle en T, on sait que :
\begin{itemize}
\item TF=6,5 cm
\item GF=9 cm
\end{itemize}

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

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

 Dans le triangle JBV rectangle en J, on sait que :
\begin{itemize}
\item JB=1,1 cm
\item JV=4,4 cm
\end{itemize}

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

 Dans le triangle FCR rectangle en F, on sait que :
\begin{itemize}
\item FR=6,8 cm
\item \(\widehat{FCR}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle SHL rectangle en S, on sait que :
\begin{itemize}
\item SH=8 cm
\item \(\widehat{HLS}\) =  14\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{GRA}\)) =  \(\dfrac{GA}{RA}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{4{,}9}{RA}\)
\par\addvspace{.5em}
 On a donc RA = \(\dfrac{4{,}9}{\sin(73^{\circ})}\) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TGF rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TFG}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TFG}\)) = \(\dfrac{TF}{GF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TFG}\)) =  \(\dfrac{6{,}5}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{TFG}\) = Arccos (\(\dfrac{6{,}5}{9}\)) \(\approx\) 44\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{JBV}\)) = \(\dfrac{JV}{JB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JBV}\)) = \(\dfrac{4{,}4}{1{,}1}\)
\par\addvspace{.5em}
 On a donc JBV = ArcTan(\(\dfrac{4{,}4}{1{,}1}\)) \(\approx\) 76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FCR}\)) = \(\dfrac{FR}{FC}\)
\par\addvspace{.5em}
tan(75\textdegree{}) = \(\dfrac{6{,}8}{FC}\)
\par\addvspace{.5em}
 On a donc FC = \(\dfrac{6{,}8}{\tan(75^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.056) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.404) {14°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SLH}\)) = \(\dfrac{SH}{SL}\)
\par\addvspace{.5em}
d'où  tan(14\textdegree{}) =  \(\dfrac{8}{SL}\)
\par\addvspace{.5em}
On a donc SL = \(\dfrac{8}{\tan(14^{\circ})}\) \(\approx\) 32,1 cm\end{document}
