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\begin{document}

\sloppy

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

 Dans le triangle CGS rectangle en C, on sait que :
\begin{itemize}
\item CS=6,3 cm
\item GS=9,3 cm
\end{itemize}

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

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

 Dans le triangle GWV rectangle en G, on sait que :
\begin{itemize}
\item GW=3,6 cm
\item \(\widehat{GWV}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle GPB rectangle en G, on sait que :
\begin{itemize}
\item GP=1,6 cm
\item PB=9,5 cm
\end{itemize}

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

 Dans le triangle NAK rectangle en N, on sait que :
\begin{itemize}
\item NK=1,7 cm
\item \(\widehat{AKN}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZDS rectangle en Z, on sait que :
\begin{itemize}
\item DS=8,7 cm
\item \(\widehat{DSZ}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{CSG}\)) = \(\dfrac{CS}{GS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{CSG}\)) =  \(\dfrac{6{,}3}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{CSG}\) = Arccos (\(\dfrac{6{,}3}{9{,}3}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{GWV}\)) = \(\dfrac{GV}{GW}\)
\par\addvspace{.5em}
d'où tan(46\textdegree{}) =\(\dfrac{GV}{3{,}6}\)
\par\addvspace{.5em}
On a donc GV = 3,6 \(\times\) tan(46\textdegree{}) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{GPB}\)) = \(\dfrac{GP}{PB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GPB}\)) = \(\dfrac{1{,}6}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{GPB}\) =ArcCos(\(\dfrac{1{,}6}{9{,}5}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NKA}\)) = \(\dfrac{NK}{AK}\)
\par\addvspace{.5em}
  d'où    cos(41\textdegree{}) =  \(\dfrac{1{,}7}{AK}\)
\par\addvspace{.5em}
 On a donc AK =  \(\dfrac{1{,}7}{\cos(41^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
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 Dans le triangle ZDS rectangle en Z, on cherche une relation entre l'angle aigu   \(\widehat{ZSD}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{ZSD}\)) = \(\dfrac{ZD}{DS}\)
\par\addvspace{.5em}
 D'où sin(22\textdegree{}) = \(\dfrac{ZD}{8{,}7}\)

On a donc ZD = 8,7 \(\times\) sin(22\textdegree{}) \(\approx\) 3,3 cm\end{document}
