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\begin{document}

\sloppy

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

 Dans le triangle SLC rectangle en S, on sait que :
\begin{itemize}
\item LC=1,5 cm
\item \(\widehat{SLC}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle MSG rectangle en M, on sait que :
\begin{itemize}
\item MS=2,7 cm
\item MG=6 cm
\end{itemize}

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

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

 Dans le triangle TWH rectangle en T, on sait que :
\begin{itemize}
\item TW=9,8 cm
\item \(\widehat{WHT}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJB rectangle en F, on sait que :
\begin{itemize}
\item FB=5,4 cm
\item JB=10 cm
\end{itemize}

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

 Dans le triangle RVA rectangle en R, on sait que :
\begin{itemize}
\item RV=0,5 cm
\item \(\widehat{VAR}\) =  40\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{SLC}\)) = \(\dfrac{SL}{LC}\)
\par\addvspace{.5em}
d'où cos(72\textdegree{}) = \(\dfrac{SL}{1{,}5}\)

On a donc SL = 1,5 \(\times\) cos(72\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MGS}\)) = \(\dfrac{MS}{MG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{MGS}\)) =  \(\dfrac{2{,}7}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MGS}\) =  ArcTan(\(\dfrac{2{,}7}{6}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle TWH rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{THW}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{THW}\)) = \(\dfrac{TW}{TH}\)
\par\addvspace{.5em}
d'où  tan(13\textdegree{}) =  \(\dfrac{9{,}8}{TH}\)
\par\addvspace{.5em}
On a donc TH = \(\dfrac{9{,}8}{\tan(13^{\circ})}\) \(\approx\) 42,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{FJB}\)) =  \(\dfrac{FB}{JB}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FJB}\)) = \(\dfrac{5{,}4}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{FJB}\) =  ArcSin(\(\dfrac{5{,}4}{10}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.542,0.761) {40°};
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 Dans le triangle RVA rectangle en R, on cherche une relation entre l'angle aigu   \(\widehat{RAV}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{RAV}\)) = \(\dfrac{RV}{VA}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{0{,}5}{VA}\)
\par\addvspace{.5em}
On a donc VA =\(\dfrac{0{,}5}{\sin(40^{\circ})}\) \(\approx\) 0,8 cm\end{document}
