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\begin{document}

\sloppy

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

 Dans le triangle RBG rectangle en R, on sait que :
\begin{itemize}
\item RB=2,7 cm
\item BG=9,2 cm
\end{itemize}

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

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

 Dans le triangle PNT rectangle en P, on sait que :
\begin{itemize}
\item PN=1 cm
\item PT=4,1 cm
\end{itemize}

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

 Dans le triangle PHS rectangle en P, on sait que :
\begin{itemize}
\item HS=2 cm
\item \(\widehat{PHS}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle WAG rectangle en W, on sait que :
\begin{itemize}
\item WG=4 cm
\item \(\widehat{WAG}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle CVW rectangle en C, on sait que :
\begin{itemize}
\item VW=7,2 cm
\item \(\widehat{CVW}\) = 53\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{RGB}\)) = \(\dfrac{RB}{BG}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RGB}\)) =\(\dfrac{2{,}7}{9{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RGB}\) = ArcSin(\(\dfrac{2{,}7}{9{,}2}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PNT}\)) = \(\dfrac{PT}{PN}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PNT}\)) = \(\dfrac{4{,}1}{1}\)
\par\addvspace{.5em}
 On a donc PNT = ArcTan(4,1) \(\approx\) 76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{PHS}\)) = \(\dfrac{PH}{HS}\)
\par\addvspace{.5em}
d'où cos(69\textdegree{}) = \(\dfrac{PH}{2}\)

On a donc PH = 2 \(\times\) cos(69\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{WAG}\)) =  \(\dfrac{WG}{AG}\)
\par\addvspace{.5em}
 d'où sin(54\textdegree{}) = \(\dfrac{4}{AG}\)
\par\addvspace{.5em}
 On a donc AG = \(\dfrac{4}{\sin(54^{\circ})}\) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.940,2.737) {53°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CVW}\)) =  \(\dfrac{CW}{VW}\)
\par\addvspace{.5em}
 d'où sin(53\textdegree{}) = \(\dfrac{CW}{7{,}2}\)

On a donc CW = 7,2 \(\times\) sin(53\textdegree{}) \(\approx\) 5,8 cm\end{document}
