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\begin{document}

\sloppy

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

 Dans le triangle JZM rectangle en J, on sait que :
\begin{itemize}
\item JZ=2,6 cm
\item \(\widehat{ZMJ}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle TLG rectangle en T, on sait que :
\begin{itemize}
\item TL=1 cm
\item LG=8,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TLG}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle RFK rectangle en R, on sait que :
\begin{itemize}
\item RF=6 cm
\item \(\widehat{FKR}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle JPS rectangle en J, on sait que :
\begin{itemize}
\item JP=2,1 cm
\item PS=8,5 cm
\end{itemize}

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

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

 Dans le triangle ABG rectangle en A, on sait que :
\begin{itemize}
\item BG=0,7 cm
\item \(\widehat{ABG}\) = 59\textdegree{}
\end{itemize}

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

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

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

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 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(34\textdegree{}) = \(\dfrac{2{,}6}{ZM}\)
\par\addvspace{.5em}
On a donc ZM =\(\dfrac{2{,}6}{\sin(34^{\circ})}\) \(\approx\) 4,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{TLG}\)) = \(\dfrac{TL}{LG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TLG}\)) = \(\dfrac{1}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{TLG}\) =ArcCos(\(\dfrac{1}{8{,}1}\)) \(\approx\)83\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{RKF}\)) = \(\dfrac{RF}{RK}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{6}{RK}\)
\par\addvspace{.5em}
On a donc RK = \(\dfrac{6}{\tan(36^{\circ})}\) \(\approx\) 8,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{JSP}\)) = \(\dfrac{JP}{PS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{JSP}\)) =\(\dfrac{2{,}1}{8{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JSP}\) = ArcSin(\(\dfrac{2{,}1}{8{,}5}\)) \(\approx\) 14\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.498,2.753) {59°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ABG}\)) = \(\dfrac{AB}{BG}\)
\par\addvspace{.5em}
d'où cos(59\textdegree{}) = \(\dfrac{AB}{0{,}7}\)

On a donc AB = 0,7 \(\times\) cos(59\textdegree{}) \(\approx\) 0,4 cm\end{document}
