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\begin{document}

\sloppy

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

 Dans le triangle WHL rectangle en W, on sait que :
\begin{itemize}
\item HL=4,5 cm
\item \(\widehat{WHL}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle DRG rectangle en D, on sait que :
\begin{itemize}
\item DR=2,1 cm
\item DG=3,9 cm
\end{itemize}

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

 Dans le triangle CPH rectangle en C, on sait que :
\begin{itemize}
\item CP=3 cm
\item \(\widehat{PHC}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle TKS rectangle en T, on sait que :
\begin{itemize}
\item TK=2,9 cm
\item TS=4,9 cm
\end{itemize}

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

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

 Dans le triangle GNW rectangle en G, on sait que :
\begin{itemize}
\item GW=8,2 cm
\item \(\widehat{GNW}\) = 56\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{WHL}\)) =  \(\dfrac{WL}{HL}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{WL}{4{,}5}\)

On a donc WL = 4,5 \(\times\) sin(50\textdegree{}) \(\approx\) 3,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{DRG}\)) = \(\dfrac{DG}{DR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DRG}\)) = \(\dfrac{3{,}9}{2{,}1}\)
\par\addvspace{.5em}
 On a donc DRG = ArcTan(\(\dfrac{3{,}9}{2{,}1}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{CHP}\)) = \(\dfrac{CP}{PH}\)
\par\addvspace{.5em}
 D'où sin(45\textdegree{}) = \(\dfrac{3}{PH}\)
\par\addvspace{.5em}
On a donc PH =\(\dfrac{3}{\sin(45^{\circ})}\) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle TKS rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TSK}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TSK}\)) = \(\dfrac{TK}{TS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TSK}\)) =  \(\dfrac{2{,}9}{4{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TSK}\) =  ArcTan(\(\dfrac{2{,}9}{4{,}9}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GNW}\)) = \(\dfrac{GW}{GN}\)
\par\addvspace{.5em}
tan(56\textdegree{}) = \(\dfrac{8{,}2}{GN}\)
\par\addvspace{.5em}
 On a donc GN = \(\dfrac{8{,}2}{\tan(56^{\circ})}\) \(\approx\) 5,5 cm\end{document}
