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\begin{document}

\sloppy

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

 Dans le triangle WSV rectangle en W, on sait que :
\begin{itemize}
\item SV=2,3 cm
\item \(\widehat{SVW}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle AJG rectangle en A, on sait que :
\begin{itemize}
\item AJ=1,2 cm
\item AG=5,7 cm
\end{itemize}

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

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

 Dans le triangle FZK rectangle en F, on sait que :
\begin{itemize}
\item FK=2,7 cm
\item \(\widehat{FZK}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle VDF rectangle en V, on sait que :
\begin{itemize}
\item DF=9,7 cm
\item \(\widehat{DFV}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle BGR rectangle en B, on sait que :
\begin{itemize}
\item BG=1,5 cm
\item BR=3,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BGR}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WVS}\)) = \(\dfrac{WS}{SV}\)
\par\addvspace{.5em}
 D'où sin(12\textdegree{}) = \(\dfrac{WS}{2{,}3}\)

On a donc WS = 2,3 \(\times\) sin(12\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{AGJ}\)) = \(\dfrac{AJ}{AG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{AGJ}\)) =  \(\dfrac{1{,}2}{5{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AGJ}\) =  ArcTan(\(\dfrac{1{,}2}{5{,}7}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FZK}\)) =  \(\dfrac{FK}{ZK}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{2{,}7}{ZK}\)
\par\addvspace{.5em}
 On a donc ZK = \(\dfrac{2{,}7}{\sin(67^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.381,0.778) {43°};
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 Dans le triangle VDF rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VFD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{VFD}\)) = \(\dfrac{VF}{DF}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{VF}{9{,}7}\)

On a donc VF = 9,7 \(\times\) cos(43\textdegree{}) \(\approx\) 7{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.724) {3,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.427,2.326) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BGR}\)) = \(\dfrac{BR}{BG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BGR}\)) = \(\dfrac{3{,}9}{1{,}5}\)
\par\addvspace{.5em}
 On a donc BGR = ArcTan(\(\dfrac{3{,}9}{1{,}5}\)) \(\approx\) 69\textdegree{}\end{document}
