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\begin{document}

\sloppy

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

 Dans le triangle GZP rectangle en G, on sait que :
\begin{itemize}
\item GZ=9 cm
\item \(\widehat{ZPG}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVR rectangle en D, on sait que :
\begin{itemize}
\item VR=4 cm
\item \(\widehat{VRD}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle FZJ rectangle en F, on sait que :
\begin{itemize}
\item FZ=1,1 cm
\item ZJ=7,5 cm
\end{itemize}

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

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

 Dans le triangle RTM rectangle en R, on sait que :
\begin{itemize}
\item RM=4 cm
\item TM=6,8 cm
\end{itemize}

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

 Dans le triangle VGH rectangle en V, on sait que :
\begin{itemize}
\item VH=0,5 cm
\item \(\widehat{VGH}\) = 73\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{GPZ}\)) = \(\dfrac{GZ}{ZP}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{9}{ZP}\)
\par\addvspace{.5em}
On a donc ZP =\(\dfrac{9}{\sin(38^{\circ})}\) \(\approx\) 14,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DRV}\)) = \(\dfrac{DR}{VR}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{DR}{4}\)

On a donc DR = 4 \(\times\) cos(27\textdegree{}) \(\approx\) 3{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{FJZ}\)) = \(\dfrac{FZ}{ZJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FJZ}\)) =\(\dfrac{1{,}1}{7{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FJZ}\) = ArcSin(\(\dfrac{1{,}1}{7{,}5}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RTM}\)) =  \(\dfrac{RM}{TM}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{RTM}\)) = \(\dfrac{4}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{RTM}\) =  ArcSin(\(\dfrac{4}{6{,}8}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{VGH}\)) = \(\dfrac{VH}{VG}\)
\par\addvspace{.5em}
tan(73\textdegree{}) = \(\dfrac{0{,}5}{VG}\)
\par\addvspace{.5em}
 On a donc VG = \(\dfrac{0{,}5}{\tan(73^{\circ})}\) \(\approx\) 0,2 cm\end{document}
