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\begin{document}

\sloppy

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

 Dans le triangle KPZ rectangle en K, on sait que :
\begin{itemize}
\item PZ=2,1 cm
\item \(\widehat{KPZ}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle HRN rectangle en H, on sait que :
\begin{itemize}
\item HN=4 cm
\item RN=8 cm
\end{itemize}

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

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

 Dans le triangle GST rectangle en G, on sait que :
\begin{itemize}
\item GS=4,6 cm
\item \(\widehat{GST}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle RNH rectangle en R, on sait que :
\begin{itemize}
\item RH=3,9 cm
\item NH=8,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RNH}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle RLC rectangle en R, on sait que :
\begin{itemize}
\item RL=8,4 cm
\item \(\widehat{RLC}\) = 57\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{KPZ}\)) = \(\dfrac{KP}{PZ}\)
\par\addvspace{.5em}
d'où cos(57\textdegree{}) = \(\dfrac{KP}{2{,}1}\)

On a donc KP = 2,1 \(\times\) cos(57\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{HNR}\)) = \(\dfrac{HN}{RN}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{HNR}\)) =  \(\dfrac{4}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{HNR}\) = Arccos (\(\dfrac{4}{8}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{GST}\)) = \(\dfrac{GS}{ST}\)
\par\addvspace{.5em}
 d'où cos(77\textdegree{}) = \(\dfrac{4{,}6}{ST}\)
\par\addvspace{.5em}
On a donc ST = \(\dfrac{4{,}6}{\cos(77^{\circ})}\)\(\approx\) 20,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RNH}\)) =  \(\dfrac{RH}{NH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{RNH}\)) = \(\dfrac{3{,}9}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{RNH}\) =  ArcSin(\(\dfrac{3{,}9}{8{,}4}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{RLC}\)) = \(\dfrac{RC}{RL}\)
\par\addvspace{.5em}
d'où tan(57\textdegree{}) =\(\dfrac{RC}{8{,}4}\)
\par\addvspace{.5em}
On a donc RC = 8,4 \(\times\) tan(57\textdegree{}) \(\approx\) 12,9 cm\end{document}
