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\begin{document}

\sloppy

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

 Dans le triangle VWT rectangle en V, on sait que :
\begin{itemize}
\item VW=0,8 cm
\item \(\widehat{WTV}\) =  39\textdegree{}
\end{itemize}

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

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

 Dans le triangle SKJ rectangle en S, on sait que :
\begin{itemize}
\item SJ=5,7 cm
\item KJ=8,7 cm
\end{itemize}

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

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

 Dans le triangle GVH rectangle en G, on sait que :
\begin{itemize}
\item VH=0,4 cm
\item \(\widehat{VHG}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle VGB rectangle en V, on sait que :
\begin{itemize}
\item VG=2 cm
\item VB=3,8 cm
\end{itemize}

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

 Dans le triangle LGM rectangle en L, on sait que :
\begin{itemize}
\item LM=2 cm
\item \(\widehat{LGM}\) = 77\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VTW}\)) = \(\dfrac{VW}{WT}\)
\par\addvspace{.5em}
 D'où sin(39\textdegree{}) = \(\dfrac{0{,}8}{WT}\)
\par\addvspace{.5em}
On a donc WT =\(\dfrac{0{,}8}{\sin(39^{\circ})}\) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SJK}\)) = \(\dfrac{SJ}{KJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SJK}\)) =  \(\dfrac{5{,}7}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SJK}\) = Arccos (\(\dfrac{5{,}7}{8{,}7}\)) \(\approx\) 49\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{GHV}\)) = \(\dfrac{GH}{VH}\)
\par\addvspace{.5em}
  d'où    cos(24\textdegree{}) =  \(\dfrac{GH}{0{,}4}\)

On a donc GH = 0,4 \(\times\) cos(24\textdegree{}) \(\approx\) 0{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VGB}\)) = \(\dfrac{VB}{VG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VGB}\)) = \(\dfrac{3{,}8}{2}\)
\par\addvspace{.5em}
 On a donc VGB = ArcTan(\(\dfrac{3{,}8}{2}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.465,1.977) {77°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LGM}\)) = \(\dfrac{LM}{LG}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{2}{LG}\)
\par\addvspace{.5em}
 On a donc LG = \(\dfrac{2}{\tan(77^{\circ})}\) \(\approx\) 0,5 cm\end{document}
