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\begin{document}

\sloppy

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

 Dans le triangle JHK rectangle en J, on sait que :
\begin{itemize}
\item JK=5,6 cm
\item HK=7,1 cm
\end{itemize}

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

 Dans le triangle MFG rectangle en M, on sait que :
\begin{itemize}
\item FG=3,2 cm
\item \(\widehat{MFG}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle CWD rectangle en C, on sait que :
\begin{itemize}
\item CW=9,6 cm
\item \(\widehat{CWD}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle GKB rectangle en G, on sait que :
\begin{itemize}
\item GK=3,2 cm
\item GB=5,5 cm
\end{itemize}

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

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

 Dans le triangle RWP rectangle en R, on sait que :
\begin{itemize}
\item WP=5,2 cm
\item \(\widehat{RWP}\) = 58\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{JHK}\)) =  \(\dfrac{JK}{HK}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{JHK}\)) = \(\dfrac{5{,}6}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{JHK}\) =  ArcSin(\(\dfrac{5{,}6}{7{,}1}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MFG}\)) = \(\dfrac{MF}{FG}\)
\par\addvspace{.5em}
d'où cos(80\textdegree{}) = \(\dfrac{MF}{3{,}2}\)

On a donc MF = 3,2 \(\times\) cos(80\textdegree{}) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{CWD}\)) = \(\dfrac{CW}{WD}\)
\par\addvspace{.5em}
 d'où cos(74\textdegree{}) = \(\dfrac{9{,}6}{WD}\)
\par\addvspace{.5em}
On a donc WD = \(\dfrac{9{,}6}{\cos(74^{\circ})}\)\(\approx\) 34,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{GBK}\)) = \(\dfrac{GK}{GB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{GBK}\)) =  \(\dfrac{3{,}2}{5{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GBK}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}5}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RWP}\)) =  \(\dfrac{RP}{WP}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{RP}{5{,}2}\)

On a donc RP = 5,2 \(\times\) sin(58\textdegree{}) \(\approx\) 4,4 cm\end{document}
