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\begin{document}

\sloppy

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

 Dans le triangle SZV rectangle en S, on sait que :
\begin{itemize}
\item SZ=7,6 cm
\item \(\widehat{SZV}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJW rectangle en F, on sait que :
\begin{itemize}
\item FW=5,4 cm
\item JW=8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{FJW}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle GWK rectangle en G, on sait que :
\begin{itemize}
\item GK=5,8 cm
\item \(\widehat{WKG}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle MLN rectangle en M, on sait que :
\begin{itemize}
\item MN=5,7 cm
\item LN=7,1 cm
\end{itemize}

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

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

 Dans le triangle VBR rectangle en V, on sait que :
\begin{itemize}
\item VR=7,9 cm
\item \(\widehat{VBR}\) = 64\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{SZV}\)) = \(\dfrac{SV}{SZ}\)
\par\addvspace{.5em}
d'où tan(52\textdegree{}) =\(\dfrac{SV}{7{,}6}\)
\par\addvspace{.5em}
On a donc SV = 7,6 \(\times\) tan(52\textdegree{}) \(\approx\) 9,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{FJW}\)) =  \(\dfrac{FW}{JW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FJW}\)) = \(\dfrac{5{,}4}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{FJW}\) =  ArcSin(\(\dfrac{5{,}4}{8}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{GKW}\)) = \(\dfrac{GW}{GK}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{GW}{5{,}8}\)
\par\addvspace{.5em}
On a donc GW =  5,8 \(\times\) tan(23\textdegree{})  \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.752,0.743) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MNL}\)) = \(\dfrac{MN}{LN}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MNL}\)) =  \(\dfrac{5{,}7}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{MNL}\) = Arccos (\(\dfrac{5{,}7}{7{,}1}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.403,2.563) {64°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VBR}\)) =  \(\dfrac{VR}{BR}\)
\par\addvspace{.5em}
 d'où sin(64\textdegree{}) = \(\dfrac{7{,}9}{BR}\)
\par\addvspace{.5em}
 On a donc BR = \(\dfrac{7{,}9}{\sin(64^{\circ})}\) \(\approx\) 8,8 cm\end{document}
