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\begin{document}

\sloppy

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

 Dans le triangle ZKF rectangle en Z, on sait que :
\begin{itemize}
\item ZF=6,2 cm
\item KF=7,1 cm
\end{itemize}

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

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

 Dans le triangle JTK rectangle en J, on sait que :
\begin{itemize}
\item TK=3,4 cm
\item \(\widehat{TKJ}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle APS rectangle en A, on sait que :
\begin{itemize}
\item PS=4,6 cm
\item \(\widehat{PSA}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle HFL rectangle en H, on sait que :
\begin{itemize}
\item HF=2,7 cm
\item FL=9 cm
\end{itemize}

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

 Dans le triangle NCZ rectangle en N, on sait que :
\begin{itemize}
\item NC=6,5 cm
\item \(\widehat{CZN}\) =  27\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{ZFK}\)) = \(\dfrac{ZF}{KF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZFK}\)) =  \(\dfrac{6{,}2}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZFK}\) = Arccos (\(\dfrac{6{,}2}{7{,}1}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{JKT}\)) = \(\dfrac{JT}{TK}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{JT}{3{,}4}\)

On a donc JT = 3,4 \(\times\) sin(36\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{ASP}\)) = \(\dfrac{AS}{PS}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{AS}{4{,}6}\)

On a donc AS = 4,6 \(\times\) cos(40\textdegree{}) \(\approx\) 3{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HFL}\)) = \(\dfrac{HF}{FL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HFL}\)) = \(\dfrac{2{,}7}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{HFL}\) =ArcCos(\(\dfrac{2{,}7}{9}\)) \(\approx\)73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NZC}\)) = \(\dfrac{NC}{CZ}\)
\par\addvspace{.5em}
 D'où sin(27\textdegree{}) = \(\dfrac{6{,}5}{CZ}\)
\par\addvspace{.5em}
On a donc CZ =\(\dfrac{6{,}5}{\sin(27^{\circ})}\) \(\approx\) 14,3 cm\end{document}
