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\begin{document}

\sloppy

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

 Dans le triangle FBN rectangle en F, on sait que :
\begin{itemize}
\item FN=9,2 cm
\item \(\widehat{FBN}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle JDG rectangle en J, on sait que :
\begin{itemize}
\item JG=4,3 cm
\item DG=7,6 cm
\end{itemize}

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

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

 Dans le triangle DRN rectangle en D, on sait que :
\begin{itemize}
\item DR=1,2 cm
\item DN=5,4 cm
\end{itemize}

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

 Dans le triangle LMH rectangle en L, on sait que :
\begin{itemize}
\item LM=4,9 cm
\item \(\widehat{MHL}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle BTV rectangle en B, on sait que :
\begin{itemize}
\item BT=4 cm
\item \(\widehat{TVB}\) =  43\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FBN}\)) = \(\dfrac{FN}{FB}\)
\par\addvspace{.5em}
tan(65\textdegree{}) = \(\dfrac{9{,}2}{FB}\)
\par\addvspace{.5em}
 On a donc FB = \(\dfrac{9{,}2}{\tan(65^{\circ})}\) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{JGD}\)) = \(\dfrac{JG}{DG}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JGD}\)) =  \(\dfrac{4{,}3}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{JGD}\) = Arccos (\(\dfrac{4{,}3}{7{,}6}\)) \(\approx\) 56\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{DRN}\)) = \(\dfrac{DN}{DR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DRN}\)) = \(\dfrac{5{,}4}{1{,}2}\)
\par\addvspace{.5em}
 On a donc DRN = ArcTan(\(\dfrac{5{,}4}{1{,}2}\)) \(\approx\) 77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{LHM}\)) = \(\dfrac{LM}{MH}\)
\par\addvspace{.5em}
 D'où sin(43\textdegree{}) = \(\dfrac{4{,}9}{MH}\)
\par\addvspace{.5em}
On a donc MH =\(\dfrac{4{,}9}{\sin(43^{\circ})}\) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BVT}\)) = \(\dfrac{BT}{BV}\)
\par\addvspace{.5em}
d'où  tan(43\textdegree{}) =  \(\dfrac{4}{BV}\)
\par\addvspace{.5em}
On a donc BV = \(\dfrac{4}{\tan(43^{\circ})}\) \(\approx\) 4,3 cm\end{document}
