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\begin{document}

\sloppy

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

 Dans le triangle FKJ rectangle en F, on sait que :
\begin{itemize}
\item FK=6,2 cm
\item \(\widehat{FKJ}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle CDJ rectangle en C, on sait que :
\begin{itemize}
\item CD=9,7 cm
\item \(\widehat{CDJ}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle FDN rectangle en F, on sait que :
\begin{itemize}
\item FD=2,5 cm
\item FN=6 cm
\end{itemize}

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

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

 Dans le triangle VTJ rectangle en V, on sait que :
\begin{itemize}
\item VT=3,1 cm
\item TJ=9,3 cm
\end{itemize}

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

 Dans le triangle RSK rectangle en R, on sait que :
\begin{itemize}
\item SK=7,6 cm
\item \(\widehat{RSK}\) = 61\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FKJ}\)) = \(\dfrac{FJ}{FK}\)
\par\addvspace{.5em}
d'où tan(50\textdegree{}) =\(\dfrac{FJ}{6{,}2}\)
\par\addvspace{.5em}
On a donc FJ = 6,2 \(\times\) tan(50\textdegree{}) \(\approx\) 7,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{CDJ}\)) = \(\dfrac{CD}{DJ}\)
\par\addvspace{.5em}
 d'où cos(72\textdegree{}) = \(\dfrac{9{,}7}{DJ}\)
\par\addvspace{.5em}
On a donc DJ = \(\dfrac{9{,}7}{\cos(72^{\circ})}\)\(\approx\) 31,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{FND}\)) = \(\dfrac{FD}{FN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FND}\)) =  \(\dfrac{2{,}5}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FND}\) =  ArcTan(\(\dfrac{2{,}5}{6}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VTJ}\)) = \(\dfrac{VT}{TJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VTJ}\)) = \(\dfrac{3{,}1}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{VTJ}\) =ArcCos(\(\dfrac{3{,}1}{9{,}3}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RSK}\)) = \(\dfrac{RS}{SK}\)
\par\addvspace{.5em}
d'où cos(61\textdegree{}) = \(\dfrac{RS}{7{,}6}\)

On a donc RS = 7,6 \(\times\) cos(61\textdegree{}) \(\approx\) 3,7 cm\end{document}
