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\begin{document}

\sloppy

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

 Dans le triangle FDA rectangle en F, on sait que :
\begin{itemize}
\item FA=2,2 cm
\item \(\widehat{FDA}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle VLW rectangle en V, on sait que :
\begin{itemize}
\item LW=4,3 cm
\item \(\widehat{VLW}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle HLW rectangle en H, on sait que :
\begin{itemize}
\item HL=4 cm
\item \(\widehat{LWH}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle RWV rectangle en R, on sait que :
\begin{itemize}
\item RW=1,4 cm
\item WV=8,1 cm
\end{itemize}

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

 Dans le triangle ACG rectangle en A, on sait que :
\begin{itemize}
\item AG=5,1 cm
\item CG=9,2 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FDA}\)) = \(\dfrac{FA}{FD}\)
\par\addvspace{.5em}
tan(75\textdegree{}) = \(\dfrac{2{,}2}{FD}\)
\par\addvspace{.5em}
 On a donc FD = \(\dfrac{2{,}2}{\tan(75^{\circ})}\) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VLW}\)) =  \(\dfrac{VW}{LW}\)
\par\addvspace{.5em}
 d'où sin(62\textdegree{}) = \(\dfrac{VW}{4{,}3}\)

On a donc VW = 4,3 \(\times\) sin(62\textdegree{}) \(\approx\) 3,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HWL}\)) = \(\dfrac{HL}{LW}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{4}{LW}\)
\par\addvspace{.5em}
On a donc LW =\(\dfrac{4}{\sin(19^{\circ})}\) \(\approx\) 12,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{RWV}\)) = \(\dfrac{RW}{WV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RWV}\)) = \(\dfrac{1{,}4}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{RWV}\) =ArcCos(\(\dfrac{1{,}4}{8{,}1}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.845,0.848) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{AGC}\)) = \(\dfrac{AG}{CG}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{AGC}\)) =  \(\dfrac{5{,}1}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{AGC}\) = Arccos (\(\dfrac{5{,}1}{9{,}2}\)) \(\approx\) 56\textdegree{} \end{document}
