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\begin{document}

\sloppy

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

 Dans le triangle SWZ rectangle en S, on sait que :
\begin{itemize}
\item SZ=4,5 cm
\item WZ=9,7 cm
\end{itemize}

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

 Dans le triangle LPV rectangle en L, on sait que :
\begin{itemize}
\item LP=2,3 cm
\item LV=6,5 cm
\end{itemize}

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

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

 Dans le triangle MTL rectangle en M, on sait que :
\begin{itemize}
\item ML=2,1 cm
\item \(\widehat{MTL}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle ACR rectangle en A, on sait que :
\begin{itemize}
\item CR=8,8 cm
\item \(\widehat{ACR}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle PWT rectangle en P, on sait que :
\begin{itemize}
\item PT=6,9 cm
\item \(\widehat{WTP}\) =  25\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.926,2.686) {?};
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 Dans le triangle SWZ rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SWZ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{SWZ}\)) =  \(\dfrac{SZ}{WZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{SWZ}\)) = \(\dfrac{4{,}5}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SWZ}\) =  ArcSin(\(\dfrac{4{,}5}{9{,}7}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LVP}\)) = \(\dfrac{LP}{LV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LVP}\)) =  \(\dfrac{2{,}3}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LVP}\) =  ArcTan(\(\dfrac{2{,}3}{6{,}5}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MTL}\)) =  \(\dfrac{ML}{TL}\)
\par\addvspace{.5em}
 d'où sin(62\textdegree{}) = \(\dfrac{2{,}1}{TL}\)
\par\addvspace{.5em}
 On a donc TL = \(\dfrac{2{,}1}{\sin(62^{\circ})}\) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ACR}\)) =  \(\dfrac{AR}{CR}\)
\par\addvspace{.5em}
 d'où sin(62\textdegree{}) = \(\dfrac{AR}{8{,}8}\)

On a donc AR = 8,8 \(\times\) sin(62\textdegree{}) \(\approx\) 7,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.299,0.935) {25°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PTW}\)) = \(\dfrac{PW}{PT}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{PW}{6{,}9}\)
\par\addvspace{.5em}
On a donc PW =  6,9 \(\times\) tan(25\textdegree{})  \(\approx\) 3,2 cm\end{document}
