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\begin{document}

\sloppy

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

 Dans le triangle SGM rectangle en S, on sait que :
\begin{itemize}
\item SG=1,2 cm
\item GM=9,3 cm
\end{itemize}

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

 Dans le triangle ANB rectangle en A, on sait que :
\begin{itemize}
\item AN=2,2 cm
\item AB=5,9 cm
\end{itemize}

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

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

 Dans le triangle PNM rectangle en P, on sait que :
\begin{itemize}
\item PN=7,3 cm
\item \(\widehat{PNM}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZNS rectangle en Z, on sait que :
\begin{itemize}
\item ZN=4,1 cm
\item \(\widehat{NSZ}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle TKL rectangle en T, on sait que :
\begin{itemize}
\item KL=8,9 cm
\item \(\widehat{TKL}\) = 58\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{SGM}\)) = \(\dfrac{SG}{GM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SGM}\)) = \(\dfrac{1{,}2}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{SGM}\) =ArcCos(\(\dfrac{1{,}2}{9{,}3}\)) \(\approx\)83\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ABN}\)) = \(\dfrac{AN}{AB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ABN}\)) =  \(\dfrac{2{,}2}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ABN}\) =  ArcTan(\(\dfrac{2{,}2}{5{,}9}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PNM}\)) = \(\dfrac{PM}{PN}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{PM}{7{,}3}\)
\par\addvspace{.5em}
On a donc PM = 7,3 \(\times\) tan(59\textdegree{}) \(\approx\) 12,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ZSN}\)) = \(\dfrac{ZN}{NS}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{4{,}1}{NS}\)
\par\addvspace{.5em}
On a donc NS =\(\dfrac{4{,}1}{\sin(41^{\circ})}\) \(\approx\) 6,2 cm
\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 TKL rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TKL}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TKL}\)) = \(\dfrac{TK}{KL}\)
\par\addvspace{.5em}
d'où cos(58\textdegree{}) = \(\dfrac{TK}{8{,}9}\)

On a donc TK = 8,9 \(\times\) cos(58\textdegree{}) \(\approx\) 4,7 cm\end{document}
