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\begin{document}

\sloppy

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

 Dans le triangle SWK rectangle en S, on sait que :
\begin{itemize}
\item SW=2,2 cm
\item \(\widehat{WKS}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle DNS rectangle en D, on sait que :
\begin{itemize}
\item DS=1,6 cm
\item \(\widehat{DNS}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFZ rectangle en D, on sait que :
\begin{itemize}
\item DZ=4 cm
\item FZ=8,3 cm
\end{itemize}

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

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

 Dans le triangle JDL rectangle en J, on sait que :
\begin{itemize}
\item JD=3,1 cm
\item DL=6,8 cm
\end{itemize}

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

 Dans le triangle LKD rectangle en L, on sait que :
\begin{itemize}
\item LD=9,8 cm
\item \(\widehat{KDL}\) =  19\textdegree{}
\end{itemize}

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

\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 (4.284,0.788) {45°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SKW}\)) = \(\dfrac{SW}{SK}\)
\par\addvspace{.5em}
d'où  tan(45\textdegree{}) =  \(\dfrac{2{,}2}{SK}\)
\par\addvspace{.5em}
On a donc SK = \(\dfrac{2{,}2}{\tan(45^{\circ})}\) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DNS}\)) =  \(\dfrac{DS}{NS}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{1{,}6}{NS}\)
\par\addvspace{.5em}
 On a donc NS = \(\dfrac{1{,}6}{\sin(67^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{DZF}\)) = \(\dfrac{DZ}{FZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DZF}\)) =  \(\dfrac{4}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{DZF}\) = Arccos (\(\dfrac{4}{8{,}3}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{JDL}\)) = \(\dfrac{JD}{DL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JDL}\)) = \(\dfrac{3{,}1}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{JDL}\) =ArcCos(\(\dfrac{3{,}1}{6{,}8}\)) \(\approx\)63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.200) {19°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LDK}\)) = \(\dfrac{LK}{LD}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{LK}{9{,}8}\)
\par\addvspace{.5em}
On a donc LK =  9,8 \(\times\) tan(19\textdegree{})  \(\approx\) 3,4 cm\end{document}
