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\begin{document}

\sloppy

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

 Dans le triangle ZAT rectangle en Z, on sait que :
\begin{itemize}
\item AT=5,4 cm
\item \(\widehat{ZAT}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZHK rectangle en Z, on sait que :
\begin{itemize}
\item ZK=4 cm
\item HK=8,9 cm
\end{itemize}

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

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

 Dans le triangle KJS rectangle en K, on sait que :
\begin{itemize}
\item JS=7 cm
\item \(\widehat{KJS}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBV rectangle en C, on sait que :
\begin{itemize}
\item CB=3,9 cm
\item \(\widehat{BVC}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle SDZ rectangle en S, on sait que :
\begin{itemize}
\item SD=1,7 cm
\item DZ=8,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{SDZ}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZAT}\)) =  \(\dfrac{ZT}{AT}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{ZT}{5{,}4}\)

On a donc ZT = 5,4 \(\times\) sin(58\textdegree{}) \(\approx\) 4,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZKH}\)) = \(\dfrac{ZK}{HK}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZKH}\)) =  \(\dfrac{4}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZKH}\) = Arccos (\(\dfrac{4}{8{,}9}\)) \(\approx\) 63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{KJS}\)) = \(\dfrac{KJ}{JS}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{KJ}{7}\)

On a donc KJ = 7 \(\times\) cos(66\textdegree{}) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CVB}\)) = \(\dfrac{CB}{BV}\)
\par\addvspace{.5em}
 D'où sin(20\textdegree{}) = \(\dfrac{3{,}9}{BV}\)
\par\addvspace{.5em}
On a donc BV =\(\dfrac{3{,}9}{\sin(20^{\circ})}\) \(\approx\) 11,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.470,1.935) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SDZ}\)) = \(\dfrac{SD}{DZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SDZ}\)) = \(\dfrac{1{,}7}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{SDZ}\) =ArcCos(\(\dfrac{1{,}7}{8{,}2}\)) \(\approx\)78\textdegree{}\end{document}
