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\begin{document}

\sloppy

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

 Dans le triangle VZN rectangle en V, on sait que :
\begin{itemize}
\item VZ=1,4 cm
\item VN=6,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{VZN}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle ZRV rectangle en Z, on sait que :
\begin{itemize}
\item RV=1,9 cm
\item \(\widehat{ZRV}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle FBN rectangle en F, on sait que :
\begin{itemize}
\item BN=2 cm
\item \(\widehat{FBN}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVM rectangle en D, on sait que :
\begin{itemize}
\item DV=3,3 cm
\item VM=8,8 cm
\end{itemize}

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

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

 Dans le triangle DFZ rectangle en D, on sait que :
\begin{itemize}
\item DZ=3,5 cm
\item \(\widehat{FZD}\) =  11\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{VZN}\)) = \(\dfrac{VN}{VZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VZN}\)) = \(\dfrac{6{,}5}{1{,}4}\)
\par\addvspace{.5em}
 On a donc VZN = ArcTan(\(\dfrac{6{,}5}{1{,}4}\)) \(\approx\) 78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZRV}\)) = \(\dfrac{ZR}{RV}\)
\par\addvspace{.5em}
d'où cos(50\textdegree{}) = \(\dfrac{ZR}{1{,}9}\)

On a donc ZR = 1,9 \(\times\) cos(50\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{FBN}\)) =  \(\dfrac{FN}{BN}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{FN}{2}\)

On a donc FN = 2 \(\times\) sin(78\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DMV}\)) = \(\dfrac{DV}{VM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DMV}\)) =\(\dfrac{3{,}3}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DMV}\) = ArcSin(\(\dfrac{3{,}3}{8{,}8}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.521) {11°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DZF}\)) = \(\dfrac{DZ}{FZ}\)
\par\addvspace{.5em}
  d'où    cos(11\textdegree{}) =  \(\dfrac{3{,}5}{FZ}\)
\par\addvspace{.5em}
 On a donc FZ =  \(\dfrac{3{,}5}{\cos(11^{\circ})}\) \(\approx\) 3,6 cm\end{document}
