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\begin{document}

\sloppy

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

 Dans le triangle VBM rectangle en V, on sait que :
\begin{itemize}
\item VM=4,3 cm
\item BM=7,4 cm
\end{itemize}

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

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

 Dans le triangle NJA rectangle en N, on sait que :
\begin{itemize}
\item JA=5,4 cm
\item \(\widehat{NJA}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle GRL rectangle en G, on sait que :
\begin{itemize}
\item GL=0,9 cm
\item \(\widehat{GRL}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle VHJ rectangle en V, on sait que :
\begin{itemize}
\item VJ=6,3 cm
\item HJ=8,5 cm
\end{itemize}

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

 Dans le triangle AGJ rectangle en A, on sait que :
\begin{itemize}
\item GJ=7,2 cm
\item \(\widehat{AGJ}\) = 51\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{VMB}\)) = \(\dfrac{VM}{BM}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VMB}\)) =  \(\dfrac{4{,}3}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{VMB}\) = Arccos (\(\dfrac{4{,}3}{7{,}4}\)) \(\approx\) 54\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NJA}\)) =  \(\dfrac{NA}{JA}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{NA}{5{,}4}\)

On a donc NA = 5,4 \(\times\) sin(78\textdegree{}) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{GRL}\)) =  \(\dfrac{GL}{RL}\)
\par\addvspace{.5em}
 d'où sin(53\textdegree{}) = \(\dfrac{0{,}9}{RL}\)
\par\addvspace{.5em}
 On a donc RL = \(\dfrac{0{,}9}{\sin(53^{\circ})}\) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VHJ}\)) =  \(\dfrac{VJ}{HJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VHJ}\)) = \(\dfrac{6{,}3}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{VHJ}\) =  ArcSin(\(\dfrac{6{,}3}{8{,}5}\)) \(\approx\) 48\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{AGJ}\)) = \(\dfrac{AG}{GJ}\)
\par\addvspace{.5em}
d'où cos(51\textdegree{}) = \(\dfrac{AG}{7{,}2}\)

On a donc AG = 7,2 \(\times\) cos(51\textdegree{}) \(\approx\) 4,5 cm\end{document}
