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\begin{document}

\sloppy

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

 Dans le triangle DKS rectangle en D, on sait que :
\begin{itemize}
\item DK=2,7 cm
\item KS=9,5 cm
\end{itemize}

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

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

 Dans le triangle RNT rectangle en R, on sait que :
\begin{itemize}
\item RN=6,8 cm
\item \(\widehat{RNT}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle CSF rectangle en C, on sait que :
\begin{itemize}
\item CS=6,1 cm
\item \(\widehat{CSF}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle ANF rectangle en A, on sait que :
\begin{itemize}
\item AN=3,1 cm
\item AF=4,9 cm
\end{itemize}

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

 Dans le triangle VAK rectangle en V, on sait que :
\begin{itemize}
\item VK=3,5 cm
\item \(\widehat{AKV}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{DSK}\)) = \(\dfrac{DK}{KS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DSK}\)) =\(\dfrac{2{,}7}{9{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DSK}\) = ArcSin(\(\dfrac{2{,}7}{9{,}5}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{RNT}\)) = \(\dfrac{RT}{RN}\)
\par\addvspace{.5em}
d'où tan(78\textdegree{}) =\(\dfrac{RT}{6{,}8}\)
\par\addvspace{.5em}
On a donc RT = 6,8 \(\times\) tan(78\textdegree{}) \(\approx\) 32 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{CSF}\)) = \(\dfrac{CS}{SF}\)
\par\addvspace{.5em}
 d'où cos(68\textdegree{}) = \(\dfrac{6{,}1}{SF}\)
\par\addvspace{.5em}
On a donc SF = \(\dfrac{6{,}1}{\cos(68^{\circ})}\)\(\approx\) 16,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{ANF}\)) = \(\dfrac{AF}{AN}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ANF}\)) = \(\dfrac{4{,}9}{3{,}1}\)
\par\addvspace{.5em}
 On a donc ANF = ArcTan(\(\dfrac{4{,}9}{3{,}1}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{VKA}\)) = \(\dfrac{VA}{VK}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{VA}{3{,}5}\)
\par\addvspace{.5em}
On a donc VA =  3,5 \(\times\) tan(22\textdegree{})  \(\approx\) 1,4 cm\end{document}
