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\begin{document}

\sloppy

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

 Dans le triangle PWV rectangle en P, on sait que :
\begin{itemize}
\item PV=9 cm
\item \(\widehat{PWV}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle RWJ rectangle en R, on sait que :
\begin{itemize}
\item RW=8,4 cm
\item \(\widehat{RWJ}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle RHC rectangle en R, on sait que :
\begin{itemize}
\item RC=4,2 cm
\item HC=8,7 cm
\end{itemize}

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

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

 Dans le triangle LMZ rectangle en L, on sait que :
\begin{itemize}
\item LZ=4,3 cm
\item MZ=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LMZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle RCZ rectangle en R, on sait que :
\begin{itemize}
\item RC=2,9 cm
\item \(\widehat{RCZ}\) = 52\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{PWV}\)) = \(\dfrac{PV}{PW}\)
\par\addvspace{.5em}
tan(46\textdegree{}) = \(\dfrac{9}{PW}\)
\par\addvspace{.5em}
 On a donc PW = \(\dfrac{9}{\tan(46^{\circ})}\) \(\approx\) 8,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{RWJ}\)) = \(\dfrac{RJ}{RW}\)
\par\addvspace{.5em}
d'où tan(69\textdegree{}) =\(\dfrac{RJ}{8{,}4}\)
\par\addvspace{.5em}
On a donc RJ = 8,4 \(\times\) tan(69\textdegree{}) \(\approx\) 21,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{RCH}\)) = \(\dfrac{RC}{HC}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{RCH}\)) =  \(\dfrac{4{,}2}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RCH}\) = Arccos (\(\dfrac{4{,}2}{8{,}7}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{LMZ}\)) =  \(\dfrac{LZ}{MZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LMZ}\)) = \(\dfrac{4{,}3}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{LMZ}\) =  ArcSin(\(\dfrac{4{,}3}{9{,}4}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{RCZ}\)) = \(\dfrac{RC}{CZ}\)
\par\addvspace{.5em}
 d'où cos(52\textdegree{}) = \(\dfrac{2{,}9}{CZ}\)
\par\addvspace{.5em}
On a donc CZ = \(\dfrac{2{,}9}{\cos(52^{\circ})}\)\(\approx\) 4,7 cm\end{document}
