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\begin{document}

\sloppy

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

 Dans le triangle HKG rectangle en H, on sait que :
\begin{itemize}
\item HK=1,8 cm
\item HG=3,5 cm
\end{itemize}

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

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

 Dans le triangle TGP rectangle en T, on sait que :
\begin{itemize}
\item GP=1,2 cm
\item \(\widehat{TGP}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle JMS rectangle en J, on sait que :
\begin{itemize}
\item JM=3,7 cm
\item \(\widehat{JMS}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle SWG rectangle en S, on sait que :
\begin{itemize}
\item SW=2 cm
\item \(\widehat{SWG}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle GTS rectangle en G, on sait que :
\begin{itemize}
\item GS=5,9 cm
\item TS=8,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GTS}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{HGK}\)) = \(\dfrac{HK}{HG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HGK}\)) =  \(\dfrac{1{,}8}{3{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HGK}\) =  ArcTan(\(\dfrac{1{,}8}{3{,}5}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{TGP}\)) = \(\dfrac{TG}{GP}\)
\par\addvspace{.5em}
d'où cos(78\textdegree{}) = \(\dfrac{TG}{1{,}2}\)

On a donc TG = 1,2 \(\times\) cos(78\textdegree{}) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{JMS}\)) = \(\dfrac{JM}{MS}\)
\par\addvspace{.5em}
 d'où cos(58\textdegree{}) = \(\dfrac{3{,}7}{MS}\)
\par\addvspace{.5em}
On a donc MS = \(\dfrac{3{,}7}{\cos(58^{\circ})}\)\(\approx\) 7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{SWG}\)) = \(\dfrac{SG}{SW}\)
\par\addvspace{.5em}
d'où tan(60\textdegree{}) =\(\dfrac{SG}{2}\)
\par\addvspace{.5em}
On a donc SG = 2 \(\times\) tan(60\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.400,2.715) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GTS}\)) =  \(\dfrac{GS}{TS}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GTS}\)) = \(\dfrac{5{,}9}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{GTS}\) =  ArcSin(\(\dfrac{5{,}9}{8{,}5}\)) \(\approx\) 44\textdegree{}\end{document}
