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\begin{document}

\sloppy

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

 Dans le triangle RSZ rectangle en R, on sait que :
\begin{itemize}
\item RZ=4,3 cm
\item SZ=7,6 cm
\end{itemize}

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

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

 Dans le triangle CRF rectangle en C, on sait que :
\begin{itemize}
\item CF=8,8 cm
\item \(\widehat{RFC}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle CAF rectangle en C, on sait que :
\begin{itemize}
\item CA=1,1 cm
\item AF=8,1 cm
\end{itemize}

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

 Dans le triangle WSC rectangle en W, on sait que :
\begin{itemize}
\item SC=9,2 cm
\item \(\widehat{SCW}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle NVS rectangle en N, on sait que :
\begin{itemize}
\item NS=6 cm
\item \(\widehat{VSN}\) =  44\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [SV]. (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 (3.667,2.000) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.871,0.844) {?};
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 Dans le triangle RSZ rectangle en R, on cherche une relation entre l'angle aigu   \(\widehat{RZS}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{RZS}\)) = \(\dfrac{RZ}{SZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{RZS}\)) =  \(\dfrac{4{,}3}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{RZS}\) = Arccos (\(\dfrac{4{,}3}{7{,}6}\)) \(\approx\) 56\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CFR}\)) = \(\dfrac{CR}{CF}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{CR}{8{,}8}\)
\par\addvspace{.5em}
On a donc CR =  8,8 \(\times\) tan(32\textdegree{})  \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CAF}\)) = \(\dfrac{CA}{AF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CAF}\)) = \(\dfrac{1{,}1}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{CAF}\) =ArcCos(\(\dfrac{1{,}1}{8{,}1}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
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 Dans le triangle WSC rectangle en W, on cherche une relation entre l'angle aigu   \(\widehat{WCS}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{WCS}\)) = \(\dfrac{WC}{SC}\)
\par\addvspace{.5em}
  d'où    cos(44\textdegree{}) =  \(\dfrac{WC}{9{,}2}\)

On a donc WC = 9,2 \(\times\) cos(44\textdegree{}) \(\approx\) 6{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{NSV}\)) = \(\dfrac{NS}{VS}\)
\par\addvspace{.5em}
  d'où    cos(44\textdegree{}) =  \(\dfrac{6}{VS}\)
\par\addvspace{.5em}
 On a donc VS =  \(\dfrac{6}{\cos(44^{\circ})}\) \(\approx\) 8,3 cm\end{document}
