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\begin{document}

\sloppy

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

 Dans le triangle DPG rectangle en D, on sait que :
\begin{itemize}
\item DG=4 cm
\item PG=9,1 cm
\end{itemize}

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

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

 Dans le triangle CGW rectangle en C, on sait que :
\begin{itemize}
\item CG=2,6 cm
\item CW=4,7 cm
\end{itemize}

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

 Dans le triangle ZWK rectangle en Z, on sait que :
\begin{itemize}
\item ZW=1,7 cm
\item \(\widehat{ZWK}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle CZD rectangle en C, on sait que :
\begin{itemize}
\item ZD=6 cm
\item \(\widehat{ZDC}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle VLG rectangle en V, on sait que :
\begin{itemize}
\item VL=6,3 cm
\item \(\widehat{VLG}\) = 80\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [VG]. (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 (3.620,0.886) {?};
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 Dans le triangle DPG rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DGP}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{DGP}\)) = \(\dfrac{DG}{PG}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DGP}\)) =  \(\dfrac{4}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{DGP}\) = Arccos (\(\dfrac{4}{9{,}1}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CGW}\)) = \(\dfrac{CW}{CG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CGW}\)) = \(\dfrac{4{,}7}{2{,}6}\)
\par\addvspace{.5em}
 On a donc CGW = ArcTan(\(\dfrac{4{,}7}{2{,}6}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{ZWK}\)) = \(\dfrac{ZW}{WK}\)
\par\addvspace{.5em}
 d'où cos(50\textdegree{}) = \(\dfrac{1{,}7}{WK}\)
\par\addvspace{.5em}
On a donc WK = \(\dfrac{1{,}7}{\cos(50^{\circ})}\)\(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CDZ}\)) = \(\dfrac{CZ}{ZD}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{CZ}{6}\)

On a donc CZ = 6 \(\times\) sin(40\textdegree{}) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{VLG}\)) = \(\dfrac{VG}{VL}\)
\par\addvspace{.5em}
d'où tan(80\textdegree{}) =\(\dfrac{VG}{6{,}3}\)
\par\addvspace{.5em}
On a donc VG = 6,3 \(\times\) tan(80\textdegree{}) \(\approx\) 35,7 cm\end{document}
