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\begin{document}

\sloppy

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

 Dans le triangle RHZ rectangle en R, on sait que :
\begin{itemize}
\item RZ=6,6 cm
\item \(\widehat{RHZ}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle RZJ rectangle en R, on sait que :
\begin{itemize}
\item ZJ=5,2 cm
\item \(\widehat{RZJ}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle MGZ rectangle en M, on sait que :
\begin{itemize}
\item MZ=3,6 cm
\item GZ=6,7 cm
\end{itemize}

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

 Dans le triangle ZLP rectangle en Z, on sait que :
\begin{itemize}
\item ZP=3,9 cm
\item LP=7,2 cm
\end{itemize}

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

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

 Dans le triangle GBS rectangle en G, on sait que :
\begin{itemize}
\item BS=4,3 cm
\item \(\widehat{BSG}\) =  36\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{RHZ}\)) =  \(\dfrac{RZ}{HZ}\)
\par\addvspace{.5em}
 d'où sin(49\textdegree{}) = \(\dfrac{6{,}6}{HZ}\)
\par\addvspace{.5em}
 On a donc HZ = \(\dfrac{6{,}6}{\sin(49^{\circ})}\) \(\approx\) 8,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RZJ}\)) = \(\dfrac{RZ}{ZJ}\)
\par\addvspace{.5em}
d'où cos(53\textdegree{}) = \(\dfrac{RZ}{5{,}2}\)

On a donc RZ = 5,2 \(\times\) cos(53\textdegree{}) \(\approx\) 3,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MGZ}\)) =  \(\dfrac{MZ}{GZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MGZ}\)) = \(\dfrac{3{,}6}{6{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{MGZ}\) =  ArcSin(\(\dfrac{3{,}6}{6{,}7}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ZPL}\)) = \(\dfrac{ZP}{LP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZPL}\)) =  \(\dfrac{3{,}9}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZPL}\) = Arccos (\(\dfrac{3{,}9}{7{,}2}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{GSB}\)) = \(\dfrac{GS}{BS}\)
\par\addvspace{.5em}
  d'où    cos(36\textdegree{}) =  \(\dfrac{GS}{4{,}3}\)

On a donc GS = 4,3 \(\times\) cos(36\textdegree{}) \(\approx\) 3{,}5 cm\end{document}
