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\begin{document}

\sloppy

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

 Dans le triangle NGW rectangle en N, on sait que :
\begin{itemize}
\item NW=4,1 cm
\item GW=8,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NGW}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle WSC rectangle en W, on sait que :
\begin{itemize}
\item WC=0,8 cm
\item \(\widehat{SCW}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle PFR rectangle en P, on sait que :
\begin{itemize}
\item PF=2,8 cm
\item FR=8,4 cm
\end{itemize}

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

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

 Dans le triangle PJS rectangle en P, on sait que :
\begin{itemize}
\item JS=0,5 cm
\item \(\widehat{PJS}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle VBT rectangle en V, on sait que :
\begin{itemize}
\item VT=2,5 cm
\item \(\widehat{VBT}\) = 72\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [TB]. (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 (3.500,0.267) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.914,2.687) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NGW}\)) =  \(\dfrac{NW}{GW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NGW}\)) = \(\dfrac{4{,}1}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{NGW}\) =  ArcSin(\(\dfrac{4{,}1}{8{,}7}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WCS}\)) = \(\dfrac{WS}{WC}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{WS}{0{,}8}\)
\par\addvspace{.5em}
On a donc WS =  0,8 \(\times\) tan(20\textdegree{})  \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{PRF}\)) = \(\dfrac{PF}{FR}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PRF}\)) =\(\dfrac{2{,}8}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PRF}\) = ArcSin(\(\dfrac{2{,}8}{8{,}4}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PJS}\)) =  \(\dfrac{PS}{JS}\)
\par\addvspace{.5em}
 d'où sin(53\textdegree{}) = \(\dfrac{PS}{0{,}5}\)

On a donc PS = 0,5 \(\times\) sin(53\textdegree{}) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.442,2.190) {72°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VBT}\)) =  \(\dfrac{VT}{BT}\)
\par\addvspace{.5em}
 d'où sin(72\textdegree{}) = \(\dfrac{2{,}5}{BT}\)
\par\addvspace{.5em}
 On a donc BT = \(\dfrac{2{,}5}{\sin(72^{\circ})}\) \(\approx\) 2,6 cm\end{document}
