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\begin{document}

\sloppy

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

 Dans le triangle MRG rectangle en M, on sait que :
\begin{itemize}
\item MR=8,4 cm
\item \(\widehat{MRG}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBL rectangle en C, on sait que :
\begin{itemize}
\item CB=1,3 cm
\item CL=4,3 cm
\end{itemize}

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

 Dans le triangle JCR rectangle en J, on sait que :
\begin{itemize}
\item JR=0,4 cm
\item \(\widehat{CRJ}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle VRW rectangle en V, on sait que :
\begin{itemize}
\item VW=5 cm
\item RW=6,7 cm
\end{itemize}

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

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

 Dans le triangle GHS rectangle en G, on sait que :
\begin{itemize}
\item HS=2 cm
\item \(\widehat{GHS}\) = 47\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{MRG}\)) = \(\dfrac{MR}{RG}\)
\par\addvspace{.5em}
 d'où cos(53\textdegree{}) = \(\dfrac{8{,}4}{RG}\)
\par\addvspace{.5em}
On a donc RG = \(\dfrac{8{,}4}{\cos(53^{\circ})}\)\(\approx\) 14 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CBL}\)) = \(\dfrac{CL}{CB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CBL}\)) = \(\dfrac{4{,}3}{1{,}3}\)
\par\addvspace{.5em}
 On a donc CBL = ArcTan(\(\dfrac{4{,}3}{1{,}3}\)) \(\approx\) 73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{JRC}\)) = \(\dfrac{JC}{JR}\)
\par\addvspace{.5em}
d'où  tan(37\textdegree{}) =  \(\dfrac{JC}{0{,}4}\)
\par\addvspace{.5em}
On a donc JC =  0,4 \(\times\) tan(37\textdegree{})  \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VWR}\)) = \(\dfrac{VW}{RW}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VWR}\)) =  \(\dfrac{5}{6{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{VWR}\) = Arccos (\(\dfrac{5}{6{,}7}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{GHS}\)) =  \(\dfrac{GS}{HS}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{GS}{2}\)

On a donc GS = 2 \(\times\) sin(47\textdegree{}) \(\approx\) 1,5 cm\end{document}
