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\begin{document}

\sloppy

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

 Dans le triangle FBV rectangle en F, on sait que :
\begin{itemize}
\item FB=0,8 cm
\item \(\widehat{BVF}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle KCZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=2,5 cm
\item \(\widehat{CZK}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAN rectangle en B, on sait que :
\begin{itemize}
\item AN=0,7 cm
\item \(\widehat{BAN}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle STA rectangle en S, on sait que :
\begin{itemize}
\item SA=5,8 cm
\item TA=7,7 cm
\end{itemize}

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

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

 Dans le triangle VLN rectangle en V, on sait que :
\begin{itemize}
\item VL=2,1 cm
\item VN=4,2 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FVB}\)) = \(\dfrac{FB}{FV}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{0{,}8}{FV}\)
\par\addvspace{.5em}
On a donc FV = \(\dfrac{0{,}8}{\tan(26^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KZC}\)) = \(\dfrac{KZ}{CZ}\)
\par\addvspace{.5em}
  d'où    cos(30\textdegree{}) =  \(\dfrac{2{,}5}{CZ}\)
\par\addvspace{.5em}
 On a donc CZ =  \(\dfrac{2{,}5}{\cos(30^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{BAN}\)) = \(\dfrac{BA}{AN}\)
\par\addvspace{.5em}
d'où cos(47\textdegree{}) = \(\dfrac{BA}{0{,}7}\)

On a donc BA = 0,7 \(\times\) cos(47\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SAT}\)) = \(\dfrac{SA}{TA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SAT}\)) =  \(\dfrac{5{,}8}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SAT}\) = Arccos (\(\dfrac{5{,}8}{7{,}7}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{VLN}\)) = \(\dfrac{VN}{VL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VLN}\)) = \(\dfrac{4{,}2}{2{,}1}\)
\par\addvspace{.5em}
 On a donc VLN = ArcTan(\(\dfrac{4{,}2}{2{,}1}\)) \(\approx\) 63\textdegree{}\end{document}
