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\begin{document}

\sloppy

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

 Dans le triangle AVB rectangle en A, on sait que :
\begin{itemize}
\item AV=2,9 cm
\item VB=9,1 cm
\end{itemize}

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

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

 Dans le triangle RFZ rectangle en R, on sait que :
\begin{itemize}
\item RF=2,6 cm
\item FZ=8,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RFZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle CBM rectangle en C, on sait que :
\begin{itemize}
\item CM=6,3 cm
\item \(\widehat{BMC}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle DWM rectangle en D, on sait que :
\begin{itemize}
\item WM=3,1 cm
\item \(\widehat{WMD}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle DBF rectangle en D, on sait que :
\begin{itemize}
\item DB=8,8 cm
\item \(\widehat{BFD}\) =  13\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ABV}\)) = \(\dfrac{AV}{VB}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ABV}\)) =\(\dfrac{2{,}9}{9{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ABV}\) = ArcSin(\(\dfrac{2{,}9}{9{,}1}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RFZ}\)) = \(\dfrac{RF}{FZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RFZ}\)) = \(\dfrac{2{,}6}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RFZ}\) =ArcCos(\(\dfrac{2{,}6}{8{,}7}\)) \(\approx\)73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CMB}\)) = \(\dfrac{CM}{BM}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{6{,}3}{BM}\)
\par\addvspace{.5em}
 On a donc BM =  \(\dfrac{6{,}3}{\cos(37^{\circ})}\) \(\approx\) 7,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DMW}\)) = \(\dfrac{DW}{WM}\)
\par\addvspace{.5em}
 D'où sin(18\textdegree{}) = \(\dfrac{DW}{3{,}1}\)

On a donc DW = 3,1 \(\times\) sin(18\textdegree{}) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.101) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.443) {13°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DFB}\)) = \(\dfrac{DB}{DF}\)
\par\addvspace{.5em}
d'où  tan(13\textdegree{}) =  \(\dfrac{8{,}8}{DF}\)
\par\addvspace{.5em}
On a donc DF = \(\dfrac{8{,}8}{\tan(13^{\circ})}\) \(\approx\) 38,1 cm\end{document}
