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\begin{document}

\sloppy

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

 Dans le triangle STK rectangle en S, on sait que :
\begin{itemize}
\item ST=1,4 cm
\item TK=9,8 cm
\end{itemize}

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

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

 Dans le triangle ZAM rectangle en Z, on sait que :
\begin{itemize}
\item ZA=6,3 cm
\item \(\widehat{AMZ}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle JZM rectangle en J, on sait que :
\begin{itemize}
\item ZM=9,2 cm
\item \(\widehat{ZMJ}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle BJZ rectangle en B, on sait que :
\begin{itemize}
\item BJ=1,9 cm
\item BZ=3,9 cm
\end{itemize}

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

 Dans le triangle JHC rectangle en J, on sait que :
\begin{itemize}
\item JH=5,1 cm
\item \(\widehat{JHC}\) = 49\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [CH]. (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 (5.285,1.627) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SKT}\)) = \(\dfrac{ST}{TK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{SKT}\)) =\(\dfrac{1{,}4}{9{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SKT}\) = ArcSin(\(\dfrac{1{,}4}{9{,}8}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ZMA}\)) = \(\dfrac{ZA}{ZM}\)
\par\addvspace{.5em}
d'où  tan(18\textdegree{}) =  \(\dfrac{6{,}3}{ZM}\)
\par\addvspace{.5em}
On a donc ZM = \(\dfrac{6{,}3}{\tan(18^{\circ})}\) \(\approx\) 19,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JMZ}\)) = \(\dfrac{JZ}{ZM}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{JZ}{9{,}2}\)

On a donc JZ = 9,2 \(\times\) sin(41\textdegree{}) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{BJZ}\)) = \(\dfrac{BZ}{BJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BJZ}\)) = \(\dfrac{3{,}9}{1{,}9}\)
\par\addvspace{.5em}
 On a donc BJZ = ArcTan(\(\dfrac{3{,}9}{1{,}9}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.389,1.933) {5,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.166,2.727) {49°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JHC}\)) = \(\dfrac{JH}{HC}\)
\par\addvspace{.5em}
 d'où cos(49\textdegree{}) = \(\dfrac{5{,}1}{HC}\)
\par\addvspace{.5em}
On a donc HC = \(\dfrac{5{,}1}{\cos(49^{\circ})}\)\(\approx\) 7,8 cm\end{document}
