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\begin{document}

\sloppy

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

 Dans le triangle TZC rectangle en T, on sait que :
\begin{itemize}
\item TC=3,6 cm
\item \(\widehat{TZC}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle DJK rectangle en D, on sait que :
\begin{itemize}
\item JK=5,3 cm
\item \(\widehat{JKD}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle JDT rectangle en J, on sait que :
\begin{itemize}
\item JD=2,9 cm
\item JT=3,9 cm
\end{itemize}

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

 Dans le triangle WZJ rectangle en W, on sait que :
\begin{itemize}
\item WJ=3,2 cm
\item \(\widehat{WZJ}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle PJA rectangle en P, on sait que :
\begin{itemize}
\item PA=4,8 cm
\item JA=9,8 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{TZC}\)) =  \(\dfrac{TC}{ZC}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{3{,}6}{ZC}\)
\par\addvspace{.5em}
 On a donc ZC = \(\dfrac{3{,}6}{\sin(78^{\circ})}\) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DKJ}\)) = \(\dfrac{DK}{JK}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{DK}{5{,}3}\)

On a donc DK = 5,3 \(\times\) cos(29\textdegree{}) \(\approx\) 4{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{JDT}\)) = \(\dfrac{JT}{JD}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JDT}\)) = \(\dfrac{3{,}9}{2{,}9}\)
\par\addvspace{.5em}
 On a donc JDT = ArcTan(\(\dfrac{3{,}9}{2{,}9}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WZJ}\)) = \(\dfrac{WJ}{WZ}\)
\par\addvspace{.5em}
tan(60\textdegree{}) = \(\dfrac{3{,}2}{WZ}\)
\par\addvspace{.5em}
 On a donc WZ = \(\dfrac{3{,}2}{\tan(60^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.711,0.870) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PAJ}\)) = \(\dfrac{PA}{JA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{PAJ}\)) =  \(\dfrac{4{,}8}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{PAJ}\) = Arccos (\(\dfrac{4{,}8}{9{,}8}\)) \(\approx\) 61\textdegree{} \end{document}
