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\begin{document}

\sloppy

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

 Dans le triangle MNG rectangle en M, on sait que :
\begin{itemize}
\item MN=2,5 cm
\item \(\widehat{NGM}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle PBH rectangle en P, on sait que :
\begin{itemize}
\item PH=4,4 cm
\item BH=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PBH}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle PCG rectangle en P, on sait que :
\begin{itemize}
\item PG=6,3 cm
\item \(\widehat{CGP}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle RJZ rectangle en R, on sait que :
\begin{itemize}
\item RZ=6,2 cm
\item JZ=9,4 cm
\end{itemize}

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

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

 Dans le triangle APN rectangle en A, on sait que :
\begin{itemize}
\item AN=4,8 cm
\item \(\widehat{APN}\) = 73\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.938,0.728) {34°};
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 Dans le triangle MNG rectangle en M, on cherche une relation entre l'angle aigu   \(\widehat{MGN}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{MGN}\)) = \(\dfrac{MN}{MG}\)
\par\addvspace{.5em}
d'où  tan(34\textdegree{}) =  \(\dfrac{2{,}5}{MG}\)
\par\addvspace{.5em}
On a donc MG = \(\dfrac{2{,}5}{\tan(34^{\circ})}\) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{PBH}\)) =  \(\dfrac{PH}{BH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PBH}\)) = \(\dfrac{4{,}4}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{PBH}\) =  ArcSin(\(\dfrac{4{,}4}{7{,}2}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{PGC}\)) = \(\dfrac{PC}{PG}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{PC}{6{,}3}\)
\par\addvspace{.5em}
On a donc PC =  6,3 \(\times\) tan(10\textdegree{})  \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{RZJ}\)) = \(\dfrac{RZ}{JZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{RZJ}\)) =  \(\dfrac{6{,}2}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{RZJ}\) = Arccos (\(\dfrac{6{,}2}{9{,}4}\)) \(\approx\) 49\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{APN}\)) =  \(\dfrac{AN}{PN}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{4{,}8}{PN}\)
\par\addvspace{.5em}
 On a donc PN = \(\dfrac{4{,}8}{\sin(73^{\circ})}\) \(\approx\) 5 cm\end{document}
