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\begin{document}

\sloppy

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

 Dans le triangle SMZ rectangle en S, on sait que :
\begin{itemize}
\item MZ=8,7 cm
\item \(\widehat{MZS}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle JNK rectangle en J, on sait que :
\begin{itemize}
\item JN=7,5 cm
\item \(\widehat{JNK}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle CPG rectangle en C, on sait que :
\begin{itemize}
\item CP=1,5 cm
\item \(\widehat{CPG}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle MBW rectangle en M, on sait que :
\begin{itemize}
\item MB=2,9 cm
\item MW=3,8 cm
\end{itemize}

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

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

 Dans le triangle KWL rectangle en K, on sait que :
\begin{itemize}
\item KW=2 cm
\item KL=5,6 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SZM}\)) = \(\dfrac{SM}{MZ}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{SM}{8{,}7}\)

On a donc SM = 8,7 \(\times\) sin(42\textdegree{}) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{JNK}\)) = \(\dfrac{JK}{JN}\)
\par\addvspace{.5em}
d'où tan(51\textdegree{}) =\(\dfrac{JK}{7{,}5}\)
\par\addvspace{.5em}
On a donc JK = 7,5 \(\times\) tan(51\textdegree{}) \(\approx\) 9,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{CPG}\)) = \(\dfrac{CP}{PG}\)
\par\addvspace{.5em}
 d'où cos(52\textdegree{}) = \(\dfrac{1{,}5}{PG}\)
\par\addvspace{.5em}
On a donc PG = \(\dfrac{1{,}5}{\cos(52^{\circ})}\)\(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MWB}\)) = \(\dfrac{MB}{MW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{MWB}\)) =  \(\dfrac{2{,}9}{3{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MWB}\) =  ArcTan(\(\dfrac{2{,}9}{3{,}8}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{KWL}\)) = \(\dfrac{KL}{KW}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KWL}\)) = \(\dfrac{5{,}6}{2}\)
\par\addvspace{.5em}
 On a donc KWL = ArcTan(\(\dfrac{5{,}6}{2}\)) \(\approx\) 70\textdegree{}\end{document}
