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\begin{document}

\sloppy

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

 Dans le triangle JGW rectangle en J, on sait que :
\begin{itemize}
\item JW=2 cm
\item \(\widehat{JGW}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFZ rectangle en T, on sait que :
\begin{itemize}
\item TZ=2,1 cm
\item \(\widehat{TFZ}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle ADT rectangle en A, on sait que :
\begin{itemize}
\item AT=3,5 cm
\item DT=9,2 cm
\end{itemize}

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

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

 Dans le triangle ZGW rectangle en Z, on sait que :
\begin{itemize}
\item ZG=2,3 cm
\item ZW=4,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ZGW}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle MTC rectangle en M, on sait que :
\begin{itemize}
\item TC=9,9 cm
\item \(\widehat{TCM}\) =  39\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{JGW}\)) = \(\dfrac{JW}{JG}\)
\par\addvspace{.5em}
tan(67\textdegree{}) = \(\dfrac{2}{JG}\)
\par\addvspace{.5em}
 On a donc JG = \(\dfrac{2}{\tan(67^{\circ})}\) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{TFZ}\)) =  \(\dfrac{TZ}{FZ}\)
\par\addvspace{.5em}
 d'où sin(65\textdegree{}) = \(\dfrac{2{,}1}{FZ}\)
\par\addvspace{.5em}
 On a donc FZ = \(\dfrac{2{,}1}{\sin(65^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{ATD}\)) = \(\dfrac{AT}{DT}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ATD}\)) =  \(\dfrac{3{,}5}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{ATD}\) = Arccos (\(\dfrac{3{,}5}{9{,}2}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{ZGW}\)) = \(\dfrac{ZW}{ZG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ZGW}\)) = \(\dfrac{4{,}4}{2{,}3}\)
\par\addvspace{.5em}
 On a donc ZGW = ArcTan(\(\dfrac{4{,}4}{2{,}3}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MCT}\)) = \(\dfrac{MC}{TC}\)
\par\addvspace{.5em}
  d'où    cos(39\textdegree{}) =  \(\dfrac{MC}{9{,}9}\)

On a donc MC = 9,9 \(\times\) cos(39\textdegree{}) \(\approx\) 7{,}7 cm\end{document}
