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\begin{document}

\sloppy

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

 Dans le triangle RKJ rectangle en R, on sait que :
\begin{itemize}
\item KJ=8,1 cm
\item \(\widehat{RKJ}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle DLW rectangle en D, on sait que :
\begin{itemize}
\item DL=1,2 cm
\item LW=7,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DLW}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MVZ rectangle en M, on sait que :
\begin{itemize}
\item MV=5,3 cm
\item \(\widehat{MVZ}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle RVF rectangle en R, on sait que :
\begin{itemize}
\item RV=8,2 cm
\item \(\widehat{VFR}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle APW rectangle en A, on sait que :
\begin{itemize}
\item AP=2 cm
\item AW=6,2 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{RKJ}\)) = \(\dfrac{RK}{KJ}\)
\par\addvspace{.5em}
d'où cos(74\textdegree{}) = \(\dfrac{RK}{8{,}1}\)

On a donc RK = 8,1 \(\times\) cos(74\textdegree{}) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DLW}\)) = \(\dfrac{DL}{LW}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DLW}\)) = \(\dfrac{1{,}2}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{DLW}\) =ArcCos(\(\dfrac{1{,}2}{7{,}9}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{MVZ}\)) = \(\dfrac{MV}{VZ}\)
\par\addvspace{.5em}
 d'où cos(59\textdegree{}) = \(\dfrac{5{,}3}{VZ}\)
\par\addvspace{.5em}
On a donc VZ = \(\dfrac{5{,}3}{\cos(59^{\circ})}\)\(\approx\) 10,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RFV}\)) = \(\dfrac{RV}{RF}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{8{,}2}{RF}\)
\par\addvspace{.5em}
On a donc RF = \(\dfrac{8{,}2}{\tan(38^{\circ})}\) \(\approx\) 10,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{AWP}\)) = \(\dfrac{AP}{AW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{AWP}\)) =  \(\dfrac{2}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AWP}\) =  ArcTan(\(\dfrac{2}{6{,}2}\)) \(\approx\) 18\textdegree{}\end{document}
