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\begin{document}

\sloppy

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

 Dans le triangle VGS rectangle en V, on sait que :
\begin{itemize}
\item GS=8,6 cm
\item \(\widehat{GSV}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle RCP rectangle en R, on sait que :
\begin{itemize}
\item RC=0,8 cm
\item \(\widehat{RCP}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle DHB rectangle en D, on sait que :
\begin{itemize}
\item DH=2,6 cm
\item HB=9,3 cm
\end{itemize}

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

 Dans le triangle VLJ rectangle en V, on sait que :
\begin{itemize}
\item VL=3,3 cm
\item LJ=9,9 cm
\end{itemize}

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

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

 Dans le triangle JHS rectangle en J, on sait que :
\begin{itemize}
\item HS=8,2 cm
\item \(\widehat{HSJ}\) =  21\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{VSG}\)) = \(\dfrac{VG}{GS}\)
\par\addvspace{.5em}
 D'où sin(44\textdegree{}) = \(\dfrac{VG}{8{,}6}\)

On a donc VG = 8,6 \(\times\) sin(44\textdegree{}) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RCP}\)) = \(\dfrac{RC}{CP}\)
\par\addvspace{.5em}
 d'où cos(63\textdegree{}) = \(\dfrac{0{,}8}{CP}\)
\par\addvspace{.5em}
On a donc CP = \(\dfrac{0{,}8}{\cos(63^{\circ})}\)\(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{DHB}\)) = \(\dfrac{DH}{HB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DHB}\)) = \(\dfrac{2{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{DHB}\) =ArcCos(\(\dfrac{2{,}6}{9{,}3}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VJL}\)) = \(\dfrac{VL}{LJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VJL}\)) =\(\dfrac{3{,}3}{9{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VJL}\) = ArcSin(\(\dfrac{3{,}3}{9{,}9}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{JSH}\)) = \(\dfrac{JS}{HS}\)
\par\addvspace{.5em}
  d'où    cos(21\textdegree{}) =  \(\dfrac{JS}{8{,}2}\)

On a donc JS = 8,2 \(\times\) cos(21\textdegree{}) \(\approx\) 7{,}7 cm\end{document}
