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\begin{document}

\sloppy

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

 Dans le triangle DSJ rectangle en D, on sait que :
\begin{itemize}
\item DS=2,7 cm
\item \(\widehat{DSJ}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle KDW rectangle en K, on sait que :
\begin{itemize}
\item KW=7,2 cm
\item \(\widehat{KDW}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTK rectangle en N, on sait que :
\begin{itemize}
\item NT=3,2 cm
\item NK=4,2 cm
\end{itemize}

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

 Dans le triangle PZW rectangle en P, on sait que :
\begin{itemize}
\item PZ=8,2 cm
\item \(\widehat{PZW}\) = 74\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 WRL rectangle en W, on sait que :
\begin{itemize}
\item WR=2,7 cm
\item WL=5,3 cm
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.413,2.465) {66°};
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 Dans le triangle DSJ rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DSJ}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{DSJ}\)) = \(\dfrac{DJ}{DS}\)
\par\addvspace{.5em}
d'où tan(66\textdegree{}) =\(\dfrac{DJ}{2{,}7}\)
\par\addvspace{.5em}
On a donc DJ = 2,7 \(\times\) tan(66\textdegree{}) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KDW}\)) = \(\dfrac{KW}{KD}\)
\par\addvspace{.5em}
tan(48\textdegree{}) = \(\dfrac{7{,}2}{KD}\)
\par\addvspace{.5em}
 On a donc KD = \(\dfrac{7{,}2}{\tan(48^{\circ})}\) \(\approx\) 6,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

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

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

cos(\(\widehat{PZW}\)) = \(\dfrac{PZ}{ZW}\)
\par\addvspace{.5em}
 d'où cos(74\textdegree{}) = \(\dfrac{8{,}2}{ZW}\)
\par\addvspace{.5em}
On a donc ZW = \(\dfrac{8{,}2}{\cos(74^{\circ})}\)\(\approx\) 29,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.419) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.302,0.841) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WLR}\)) = \(\dfrac{WR}{WL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WLR}\)) =  \(\dfrac{2{,}7}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WLR}\) =  ArcTan(\(\dfrac{2{,}7}{5{,}3}\)) \(\approx\) 27\textdegree{}\end{document}
