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\begin{document}

\sloppy

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

 Dans le triangle HGP rectangle en H, on sait que :
\begin{itemize}
\item HG=3 cm
\item HP=6,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HGP}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle ZCH rectangle en Z, on sait que :
\begin{itemize}
\item ZC=1,9 cm
\item \(\widehat{CHZ}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRB rectangle en T, on sait que :
\begin{itemize}
\item TB=5,5 cm
\item RB=9 cm
\end{itemize}

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

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

 Dans le triangle MCS rectangle en M, on sait que :
\begin{itemize}
\item MC=2 cm
\item \(\widehat{MCS}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle RGF rectangle en R, on sait que :
\begin{itemize}
\item RF=1 cm
\item \(\widehat{GFR}\) =  21\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{HGP}\)) = \(\dfrac{HP}{HG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HGP}\)) = \(\dfrac{6{,}6}{3}\)
\par\addvspace{.5em}
 On a donc HGP = ArcTan(\(\dfrac{6{,}6}{3}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZHC}\)) = \(\dfrac{ZC}{CH}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{1{,}9}{CH}\)
\par\addvspace{.5em}
On a donc CH =\(\dfrac{1{,}9}{\sin(19^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


 Dans le triangle TRB rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TBR}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TBR}\)) = \(\dfrac{TB}{RB}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TBR}\)) =  \(\dfrac{5{,}5}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{TBR}\) = Arccos (\(\dfrac{5{,}5}{9}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MCS}\)) = \(\dfrac{MS}{MC}\)
\par\addvspace{.5em}
d'où tan(66\textdegree{}) =\(\dfrac{MS}{2}\)
\par\addvspace{.5em}
On a donc MS = 2 \(\times\) tan(66\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RFG}\)) = \(\dfrac{RG}{RF}\)
\par\addvspace{.5em}
d'où  tan(21\textdegree{}) =  \(\dfrac{RG}{1}\)
\par\addvspace{.5em}
On a donc RG =  1 \(\times\) tan(21\textdegree{})  \(\approx\) 0,4 cm\end{document}
