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\begin{document}

\sloppy

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

 Dans le triangle MZS rectangle en M, on sait que :
\begin{itemize}
\item MZ=1 cm
\item MS=6,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MZS}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle VZR rectangle en V, on sait que :
\begin{itemize}
\item VZ=3 cm
\item ZR=7,9 cm
\end{itemize}

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

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

 Dans le triangle LSW rectangle en L, on sait que :
\begin{itemize}
\item LS=5,1 cm
\item \(\widehat{SWL}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle BML rectangle en B, on sait que :
\begin{itemize}
\item BL=5 cm
\item \(\widehat{MLB}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle DPS rectangle en D, on sait que :
\begin{itemize}
\item DP=5,4 cm
\item \(\widehat{DPS}\) = 71\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{MZS}\)) = \(\dfrac{MS}{MZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MZS}\)) = \(\dfrac{6{,}6}{1}\)
\par\addvspace{.5em}
 On a donc MZS = ArcTan(6,6) \(\approx\) 81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VRZ}\)) = \(\dfrac{VZ}{ZR}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VRZ}\)) =\(\dfrac{3}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VRZ}\) = ArcSin(\(\dfrac{3}{7{,}9}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LWS}\)) = \(\dfrac{LS}{SW}\)
\par\addvspace{.5em}
 D'où sin(30\textdegree{}) = \(\dfrac{5{,}1}{SW}\)
\par\addvspace{.5em}
On a donc SW =\(\dfrac{5{,}1}{\sin(30^{\circ})}\) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{BLM}\)) = \(\dfrac{BM}{BL}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{BM}{5}\)
\par\addvspace{.5em}
On a donc BM =  5 \(\times\) tan(38\textdegree{})  \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.823) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.437,2.234) {71°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DPS}\)) = \(\dfrac{DS}{DP}\)
\par\addvspace{.5em}
d'où tan(71\textdegree{}) =\(\dfrac{DS}{5{,}4}\)
\par\addvspace{.5em}
On a donc DS = 5,4 \(\times\) tan(71\textdegree{}) \(\approx\) 15,7 cm\end{document}
