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\begin{document}

\sloppy

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

 Dans le triangle SBV rectangle en S, on sait que :
\begin{itemize}
\item SV=4,7 cm
\item \(\widehat{SBV}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle AGM rectangle en A, on sait que :
\begin{itemize}
\item AG=1,3 cm
\item GM=9,9 cm
\end{itemize}

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

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

 Dans le triangle MWT rectangle en M, on sait que :
\begin{itemize}
\item MW=1,8 cm
\item WT=9,7 cm
\end{itemize}

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

 Dans le triangle HCG rectangle en H, on sait que :
\begin{itemize}
\item HC=6 cm
\item \(\widehat{HCG}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle JMN rectangle en J, on sait que :
\begin{itemize}
\item MN=2,8 cm
\item \(\widehat{JMN}\) = 63\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{SBV}\)) = \(\dfrac{SV}{SB}\)
\par\addvspace{.5em}
tan(65\textdegree{}) = \(\dfrac{4{,}7}{SB}\)
\par\addvspace{.5em}
 On a donc SB = \(\dfrac{4{,}7}{\tan(65^{\circ})}\) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{AMG}\)) = \(\dfrac{AG}{GM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{AMG}\)) =\(\dfrac{1{,}3}{9{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AMG}\) = ArcSin(\(\dfrac{1{,}3}{9{,}9}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MWT}\)) = \(\dfrac{MW}{WT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MWT}\)) = \(\dfrac{1{,}8}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{MWT}\) =ArcCos(\(\dfrac{1{,}8}{9{,}7}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HCG}\)) = \(\dfrac{HC}{CG}\)
\par\addvspace{.5em}
 d'où cos(60\textdegree{}) = \(\dfrac{6}{CG}\)
\par\addvspace{.5em}
On a donc CG = \(\dfrac{6}{\cos(60^{\circ})}\)\(\approx\) 12 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.418) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.398,2.613) {63°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JMN}\)) =  \(\dfrac{JN}{MN}\)
\par\addvspace{.5em}
 d'où sin(63\textdegree{}) = \(\dfrac{JN}{2{,}8}\)

On a donc JN = 2,8 \(\times\) sin(63\textdegree{}) \(\approx\) 2,5 cm\end{document}
