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\begin{document}

\sloppy

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

 Dans le triangle LAW rectangle en L, on sait que :
\begin{itemize}
\item LW=2,3 cm
\item \(\widehat{AWL}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle PSA rectangle en P, on sait que :
\begin{itemize}
\item PS=3 cm
\item PA=5,1 cm
\end{itemize}

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

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

 Dans le triangle KNZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=9,1 cm
\item \(\widehat{NZK}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle JFD rectangle en J, on sait que :
\begin{itemize}
\item FD=7,7 cm
\item \(\widehat{FDJ}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle HVZ rectangle en H, on sait que :
\begin{itemize}
\item HZ=5,3 cm
\item VZ=9,7 cm
\end{itemize}

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

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

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

cos(\(\widehat{LWA}\)) = \(\dfrac{LW}{AW}\)
\par\addvspace{.5em}
  d'où    cos(32\textdegree{}) =  \(\dfrac{2{,}3}{AW}\)
\par\addvspace{.5em}
 On a donc AW =  \(\dfrac{2{,}3}{\cos(32^{\circ})}\) \(\approx\) 2,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PAS}\)) = \(\dfrac{PS}{PA}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{PAS}\)) =  \(\dfrac{3}{5{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PAS}\) =  ArcTan(\(\dfrac{3}{5{,}1}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{KZN}\)) = \(\dfrac{KN}{KZ}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{KN}{9{,}1}\)
\par\addvspace{.5em}
On a donc KN =  9,1 \(\times\) tan(40\textdegree{})  \(\approx\) 7,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{JDF}\)) = \(\dfrac{JD}{FD}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{JD}{7{,}7}\)

On a donc JD = 7,7 \(\times\) cos(35\textdegree{}) \(\approx\) 6{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{HVZ}\)) =  \(\dfrac{HZ}{VZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HVZ}\)) = \(\dfrac{5{,}3}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{HVZ}\) =  ArcSin(\(\dfrac{5{,}3}{9{,}7}\)) \(\approx\) 33\textdegree{}\end{document}
