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\begin{document}

\sloppy

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

 Dans le triangle BSP rectangle en B, on sait que :
\begin{itemize}
\item BS=1,2 cm
\item SP=8,3 cm
\end{itemize}

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

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

 Dans le triangle WBN rectangle en W, on sait que :
\begin{itemize}
\item BN=1 cm
\item \(\widehat{BNW}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle MKN rectangle en M, on sait que :
\begin{itemize}
\item MK=9,1 cm
\item \(\widehat{MKN}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle JNG rectangle en J, on sait que :
\begin{itemize}
\item JN=3,3 cm
\item \(\widehat{NGJ}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle MCV rectangle en M, on sait que :
\begin{itemize}
\item MC=2 cm
\item CV=8,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MCV}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BPS}\)) = \(\dfrac{BS}{SP}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BPS}\)) =\(\dfrac{1{,}2}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BPS}\) = ArcSin(\(\dfrac{1{,}2}{8{,}3}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{WNB}\)) = \(\dfrac{WB}{BN}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{WB}{1}\)

On a donc WB = 1 \(\times\) sin(42\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MKN}\)) = \(\dfrac{MK}{KN}\)
\par\addvspace{.5em}
 d'où cos(51\textdegree{}) = \(\dfrac{9{,}1}{KN}\)
\par\addvspace{.5em}
On a donc KN = \(\dfrac{9{,}1}{\cos(51^{\circ})}\)\(\approx\) 14,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{JGN}\)) = \(\dfrac{JN}{JG}\)
\par\addvspace{.5em}
d'où  tan(17\textdegree{}) =  \(\dfrac{3{,}3}{JG}\)
\par\addvspace{.5em}
On a donc JG = \(\dfrac{3{,}3}{\tan(17^{\circ})}\) \(\approx\) 10,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{MCV}\)) = \(\dfrac{MC}{CV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MCV}\)) = \(\dfrac{2}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{MCV}\) =ArcCos(\(\dfrac{2}{8{,}1}\)) \(\approx\)76\textdegree{}\end{document}
