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\begin{document}

\sloppy

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

 Dans le triangle BVG rectangle en B, on sait que :
\begin{itemize}
\item BV=1 cm
\item BG=6 cm
\end{itemize}

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

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

 Dans le triangle GVT rectangle en G, on sait que :
\begin{itemize}
\item GT=3,5 cm
\item \(\widehat{VTG}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle WVC rectangle en W, on sait que :
\begin{itemize}
\item VC=2,8 cm
\item \(\widehat{WVC}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle VKG rectangle en V, on sait que :
\begin{itemize}
\item VG=6 cm
\item KG=9,1 cm
\end{itemize}

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

 Dans le triangle VSA rectangle en V, on sait que :
\begin{itemize}
\item VS=8,9 cm
\item \(\widehat{VSA}\) = 51\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{BGV}\)) = \(\dfrac{BV}{BG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BGV}\)) =  \(\dfrac{1}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BGV}\) =  ArcTan(\(\dfrac{1}{6}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{GTV}\)) = \(\dfrac{GV}{GT}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{GV}{3{,}5}\)
\par\addvspace{.5em}
On a donc GV =  3,5 \(\times\) tan(33\textdegree{})  \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{WVC}\)) =  \(\dfrac{WC}{VC}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{WC}{2{,}8}\)

On a donc WC = 2,8 \(\times\) sin(57\textdegree{}) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VKG}\)) =  \(\dfrac{VG}{KG}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VKG}\)) = \(\dfrac{6}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{VKG}\) =  ArcSin(\(\dfrac{6}{9{,}1}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{VSA}\)) = \(\dfrac{VS}{SA}\)
\par\addvspace{.5em}
 d'où cos(51\textdegree{}) = \(\dfrac{8{,}9}{SA}\)
\par\addvspace{.5em}
On a donc SA = \(\dfrac{8{,}9}{\cos(51^{\circ})}\)\(\approx\) 14,1 cm\end{document}
