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\begin{document}

\sloppy

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

 Dans le triangle WVZ rectangle en W, on sait que :
\begin{itemize}
\item WV=2,3 cm
\item WZ=4,6 cm
\end{itemize}

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

 Dans le triangle MNL rectangle en M, on sait que :
\begin{itemize}
\item ML=4,9 cm
\item \(\widehat{NLM}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle HVK rectangle en H, on sait que :
\begin{itemize}
\item VK=3,6 cm
\item \(\widehat{HVK}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle GVP rectangle en G, on sait que :
\begin{itemize}
\item VP=2 cm
\item \(\widehat{VPG}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle RJC rectangle en R, on sait que :
\begin{itemize}
\item RJ=1,6 cm
\item RC=6,3 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{WVZ}\)) = \(\dfrac{WZ}{WV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WVZ}\)) = \(\dfrac{4{,}6}{2{,}3}\)
\par\addvspace{.5em}
 On a donc WVZ = ArcTan(\(\dfrac{4{,}6}{2{,}3}\)) \(\approx\) 63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MLN}\)) = \(\dfrac{ML}{NL}\)
\par\addvspace{.5em}
  d'où    cos(32\textdegree{}) =  \(\dfrac{4{,}9}{NL}\)
\par\addvspace{.5em}
 On a donc NL =  \(\dfrac{4{,}9}{\cos(32^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HVK}\)) =  \(\dfrac{HK}{VK}\)
\par\addvspace{.5em}
 d'où sin(80\textdegree{}) = \(\dfrac{HK}{3{,}6}\)

On a donc HK = 3,6 \(\times\) sin(80\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GPV}\)) = \(\dfrac{GV}{VP}\)
\par\addvspace{.5em}
 D'où sin(15\textdegree{}) = \(\dfrac{GV}{2}\)

On a donc GV = 2 \(\times\) sin(15\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.044) {6,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.394) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RCJ}\)) = \(\dfrac{RJ}{RC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RCJ}\)) =  \(\dfrac{1{,}6}{6{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RCJ}\) =  ArcTan(\(\dfrac{1{,}6}{6{,}3}\)) \(\approx\) 14\textdegree{}\end{document}
