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\begin{document}

\sloppy

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

 Dans le triangle HAF rectangle en H, on sait que :
\begin{itemize}
\item HF=5,7 cm
\item \(\widehat{AFH}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle RSJ rectangle en R, on sait que :
\begin{itemize}
\item RJ=5,2 cm
\item SJ=7 cm
\end{itemize}

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

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

 Dans le triangle PVB rectangle en P, on sait que :
\begin{itemize}
\item VB=4,2 cm
\item \(\widehat{VBP}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle VBN rectangle en V, on sait que :
\begin{itemize}
\item BN=5,5 cm
\item \(\widehat{BNV}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle MVB rectangle en M, on sait que :
\begin{itemize}
\item MV=1,6 cm
\item MB=6,4 cm
\end{itemize}

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

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

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

cos(\(\widehat{HFA}\)) = \(\dfrac{HF}{AF}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{5{,}7}{AF}\)
\par\addvspace{.5em}
 On a donc AF =  \(\dfrac{5{,}7}{\cos(43^{\circ})}\) \(\approx\) 7,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RJS}\)) = \(\dfrac{RJ}{SJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{RJS}\)) =  \(\dfrac{5{,}2}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RJS}\) = Arccos (\(\dfrac{5{,}2}{7}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{PBV}\)) = \(\dfrac{PV}{VB}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{PV}{4{,}2}\)

On a donc PV = 4,2 \(\times\) sin(38\textdegree{}) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VNB}\)) = \(\dfrac{VN}{BN}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{VN}{5{,}5}\)

On a donc VN = 5,5 \(\times\) cos(29\textdegree{}) \(\approx\) 4{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{MVB}\)) = \(\dfrac{MB}{MV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MVB}\)) = \(\dfrac{6{,}4}{1{,}6}\)
\par\addvspace{.5em}
 On a donc MVB = ArcTan(\(\dfrac{6{,}4}{1{,}6}\)) \(\approx\) 76\textdegree{}\end{document}
