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\begin{document}

\sloppy

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

 Dans le triangle SFB rectangle en S, on sait que :
\begin{itemize}
\item SB=5,9 cm
\item \(\widehat{SFB}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle DLA rectangle en D, on sait que :
\begin{itemize}
\item DL=2,7 cm
\item LA=8,2 cm
\end{itemize}

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

 Dans le triangle VMT rectangle en V, on sait que :
\begin{itemize}
\item VM=2,3 cm
\item VT=6,4 cm
\end{itemize}

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

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

 Dans le triangle VAS rectangle en V, on sait que :
\begin{itemize}
\item AS=2,2 cm
\item \(\widehat{ASV}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVK rectangle en D, on sait que :
\begin{itemize}
\item DK=4,9 cm
\item \(\widehat{VKD}\) =  15\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SFB}\)) =  \(\dfrac{SB}{FB}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{5{,}9}{FB}\)
\par\addvspace{.5em}
 On a donc FB = \(\dfrac{5{,}9}{\sin(75^{\circ})}\) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DLA}\)) = \(\dfrac{DL}{LA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DLA}\)) = \(\dfrac{2{,}7}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{DLA}\) =ArcCos(\(\dfrac{2{,}7}{8{,}2}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VTM}\)) = \(\dfrac{VM}{VT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{VTM}\)) =  \(\dfrac{2{,}3}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VTM}\) =  ArcTan(\(\dfrac{2{,}3}{6{,}4}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VSA}\)) = \(\dfrac{VS}{AS}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{VS}{2{,}2}\)

On a donc VS = 2,2 \(\times\) cos(22\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DKV}\)) = \(\dfrac{DV}{DK}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{DV}{4{,}9}\)
\par\addvspace{.5em}
On a donc DV =  4,9 \(\times\) tan(15\textdegree{})  \(\approx\) 1,3 cm\end{document}
