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\begin{document}

\sloppy

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

 Dans le triangle BWH rectangle en B, on sait que :
\begin{itemize}
\item BH=10 cm
\item \(\widehat{WHB}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle PZA rectangle en P, on sait que :
\begin{itemize}
\item PZ=2,1 cm
\item PA=6,1 cm
\end{itemize}

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

 Dans le triangle MPG rectangle en M, on sait que :
\begin{itemize}
\item PG=9,3 cm
\item \(\widehat{PGM}\) =  39\textdegree{}
\end{itemize}

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

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

 Dans le triangle VGW rectangle en V, on sait que :
\begin{itemize}
\item VG=7,1 cm
\item \(\widehat{GWV}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZRG rectangle en Z, on sait que :
\begin{itemize}
\item ZR=2,9 cm
\item ZG=4,6 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{BHW}\)) = \(\dfrac{BH}{WH}\)
\par\addvspace{.5em}
  d'où    cos(28\textdegree{}) =  \(\dfrac{10}{WH}\)
\par\addvspace{.5em}
 On a donc WH =  \(\dfrac{10}{\cos(28^{\circ})}\) \(\approx\) 11,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PZA}\)) = \(\dfrac{PA}{PZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PZA}\)) = \(\dfrac{6{,}1}{2{,}1}\)
\par\addvspace{.5em}
 On a donc PZA = ArcTan(\(\dfrac{6{,}1}{2{,}1}\)) \(\approx\) 71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MGP}\)) = \(\dfrac{MP}{PG}\)
\par\addvspace{.5em}
 D'où sin(39\textdegree{}) = \(\dfrac{MP}{9{,}3}\)

On a donc MP = 9,3 \(\times\) sin(39\textdegree{}) \(\approx\) 5,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VWG}\)) = \(\dfrac{VG}{VW}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{7{,}1}{VW}\)
\par\addvspace{.5em}
On a donc VW = \(\dfrac{7{,}1}{\tan(36^{\circ})}\) \(\approx\) 9,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.080,0.718) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZGR}\)) = \(\dfrac{ZR}{ZG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZGR}\)) =  \(\dfrac{2{,}9}{4{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZGR}\) =  ArcTan(\(\dfrac{2{,}9}{4{,}6}\)) \(\approx\) 32\textdegree{}\end{document}
