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\begin{document}

\sloppy

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

 Dans le triangle JVN rectangle en J, on sait que :
\begin{itemize}
\item JN=5,9 cm
\item VN=9,5 cm
\end{itemize}

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

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

 Dans le triangle SKM rectangle en S, on sait que :
\begin{itemize}
\item KM=9,9 cm
\item \(\widehat{SKM}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVP rectangle en D, on sait que :
\begin{itemize}
\item DP=4,2 cm
\item VP=7,2 cm
\end{itemize}

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

 Dans le triangle DPS rectangle en D, on sait que :
\begin{itemize}
\item PS=2,2 cm
\item \(\widehat{DPS}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle HFS rectangle en H, on sait que :
\begin{itemize}
\item HF=9,8 cm
\item \(\widehat{FSH}\) =  28\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{JNV}\)) = \(\dfrac{JN}{VN}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JNV}\)) =  \(\dfrac{5{,}9}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{JNV}\) = Arccos (\(\dfrac{5{,}9}{9{,}5}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SKM}\)) = \(\dfrac{SK}{KM}\)
\par\addvspace{.5em}
d'où cos(75\textdegree{}) = \(\dfrac{SK}{9{,}9}\)

On a donc SK = 9,9 \(\times\) cos(75\textdegree{}) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{DVP}\)) =  \(\dfrac{DP}{VP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DVP}\)) = \(\dfrac{4{,}2}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{DVP}\) =  ArcSin(\(\dfrac{4{,}2}{7{,}2}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DPS}\)) =  \(\dfrac{DS}{PS}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{DS}{2{,}2}\)

On a donc DS = 2,2 \(\times\) sin(68\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{HSF}\)) = \(\dfrac{HF}{FS}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{9{,}8}{FS}\)
\par\addvspace{.5em}
On a donc FS =\(\dfrac{9{,}8}{\sin(28^{\circ})}\) \(\approx\) 20,9 cm\end{document}
