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\begin{document}

\sloppy

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

 Dans le triangle FJV rectangle en F, on sait que :
\begin{itemize}
\item FV=9,6 cm
\item \(\widehat{JVF}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle BWT rectangle en B, on sait que :
\begin{itemize}
\item BT=3,8 cm
\item \(\widehat{BWT}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle CFW rectangle en C, on sait que :
\begin{itemize}
\item CF=3,2 cm
\item FW=9,2 cm
\end{itemize}

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

 Dans le triangle VHB rectangle en V, on sait que :
\begin{itemize}
\item VH=2,1 cm
\item \(\widehat{VHB}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle RCG rectangle en R, on sait que :
\begin{itemize}
\item RC=3,2 cm
\item RG=5,3 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FVJ}\)) = \(\dfrac{FJ}{FV}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{FJ}{9{,}6}\)
\par\addvspace{.5em}
On a donc FJ =  9,6 \(\times\) tan(33\textdegree{})  \(\approx\) 6,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{BWT}\)) =  \(\dfrac{BT}{WT}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{3{,}8}{WT}\)
\par\addvspace{.5em}
 On a donc WT = \(\dfrac{3{,}8}{\sin(78^{\circ})}\) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CFW}\)) = \(\dfrac{CF}{FW}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CFW}\)) = \(\dfrac{3{,}2}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{CFW}\) =ArcCos(\(\dfrac{3{,}2}{9{,}2}\)) \(\approx\)70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VHB}\)) = \(\dfrac{VB}{VH}\)
\par\addvspace{.5em}
d'où tan(57\textdegree{}) =\(\dfrac{VB}{2{,}1}\)
\par\addvspace{.5em}
On a donc VB = 2,1 \(\times\) tan(57\textdegree{}) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.681,1.933) {3,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.177,0.712) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RGC}\)) = \(\dfrac{RC}{RG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RGC}\)) =  \(\dfrac{3{,}2}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RGC}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}3}\)) \(\approx\) 31\textdegree{}\end{document}
