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\begin{document}

\sloppy

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

 Dans le triangle PHD rectangle en P, on sait que :
\begin{itemize}
\item HD=8,2 cm
\item \(\widehat{HDP}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSR rectangle en K, on sait que :
\begin{itemize}
\item KR=1,2 cm
\item \(\widehat{KSR}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle HZP rectangle en H, on sait que :
\begin{itemize}
\item ZP=3,8 cm
\item \(\widehat{HZP}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle JGT rectangle en J, on sait que :
\begin{itemize}
\item JG=2,2 cm
\item JT=6,2 cm
\end{itemize}

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

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

 Dans le triangle AMJ rectangle en A, on sait que :
\begin{itemize}
\item AM=3 cm
\item AJ=3,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AMJ}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

cos(\(\widehat{PDH}\)) = \(\dfrac{PD}{HD}\)
\par\addvspace{.5em}
  d'où    cos(18\textdegree{}) =  \(\dfrac{PD}{8{,}2}\)

On a donc PD = 8,2 \(\times\) cos(18\textdegree{}) \(\approx\) 7{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KSR}\)) =  \(\dfrac{KR}{SR}\)
\par\addvspace{.5em}
 d'où sin(46\textdegree{}) = \(\dfrac{1{,}2}{SR}\)
\par\addvspace{.5em}
 On a donc SR = \(\dfrac{1{,}2}{\sin(46^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{HZP}\)) = \(\dfrac{HZ}{ZP}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{HZ}{3{,}8}\)

On a donc HZ = 3,8 \(\times\) cos(66\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{JTG}\)) = \(\dfrac{JG}{JT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{JTG}\)) =  \(\dfrac{2{,}2}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JTG}\) =  ArcTan(\(\dfrac{2{,}2}{6{,}2}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.145,2.728) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AMJ}\)) = \(\dfrac{AJ}{AM}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{AMJ}\)) = \(\dfrac{3{,}5}{3}\)
\par\addvspace{.5em}
 On a donc AMJ = ArcTan(\(\dfrac{3{,}5}{3}\)) \(\approx\) 49\textdegree{}\end{document}
