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\begin{document}

\sloppy

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

 Dans le triangle JBR rectangle en J, on sait que :
\begin{itemize}
\item JR=4,2 cm
\item BR=7,8 cm
\end{itemize}

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

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

 Dans le triangle PVH rectangle en P, on sait que :
\begin{itemize}
\item PV=6 cm
\item \(\widehat{PVH}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle VNZ rectangle en V, on sait que :
\begin{itemize}
\item NZ=7,5 cm
\item \(\widehat{NZV}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWZ rectangle en F, on sait que :
\begin{itemize}
\item FZ=8,9 cm
\item \(\widehat{WZF}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle KPN rectangle en K, on sait que :
\begin{itemize}
\item KP=2,9 cm
\item KN=5,3 cm
\end{itemize}

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

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

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

cos(\(\widehat{JRB}\)) = \(\dfrac{JR}{BR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JRB}\)) =  \(\dfrac{4{,}2}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{JRB}\) = Arccos (\(\dfrac{4{,}2}{7{,}8}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PVH}\)) = \(\dfrac{PV}{VH}\)
\par\addvspace{.5em}
 d'où cos(55\textdegree{}) = \(\dfrac{6}{VH}\)
\par\addvspace{.5em}
On a donc VH = \(\dfrac{6}{\cos(55^{\circ})}\)\(\approx\) 10,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{VZN}\)) = \(\dfrac{VZ}{NZ}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{VZ}{7{,}5}\)

On a donc VZ = 7,5 \(\times\) cos(38\textdegree{}) \(\approx\) 5{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FZW}\)) = \(\dfrac{FW}{FZ}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{FW}{8{,}9}\)
\par\addvspace{.5em}
On a donc FW =  8,9 \(\times\) tan(33\textdegree{})  \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{KPN}\)) = \(\dfrac{KN}{KP}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KPN}\)) = \(\dfrac{5{,}3}{2{,}9}\)
\par\addvspace{.5em}
 On a donc KPN = ArcTan(\(\dfrac{5{,}3}{2{,}9}\)) \(\approx\) 61\textdegree{}\end{document}
