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\begin{document}

\sloppy

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

 Dans le triangle LAD rectangle en L, on sait que :
\begin{itemize}
\item LD=6,5 cm
\item \(\widehat{LAD}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle KVB rectangle en K, on sait que :
\begin{itemize}
\item KB=1,2 cm
\item \(\widehat{KVB}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWP rectangle en F, on sait que :
\begin{itemize}
\item WP=9,7 cm
\item \(\widehat{FWP}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle VJP rectangle en V, on sait que :
\begin{itemize}
\item VP=4,8 cm
\item JP=7,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{VJP}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle BFD rectangle en B, on sait que :
\begin{itemize}
\item BF=1,6 cm
\item FD=8,8 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LAD}\)) =  \(\dfrac{LD}{AD}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{6{,}5}{AD}\)
\par\addvspace{.5em}
 On a donc AD = \(\dfrac{6{,}5}{\sin(68^{\circ})}\) \(\approx\) 7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KVB}\)) = \(\dfrac{KB}{KV}\)
\par\addvspace{.5em}
tan(54\textdegree{}) = \(\dfrac{1{,}2}{KV}\)
\par\addvspace{.5em}
 On a donc KV = \(\dfrac{1{,}2}{\tan(54^{\circ})}\) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FWP}\)) =  \(\dfrac{FP}{WP}\)
\par\addvspace{.5em}
 d'où sin(60\textdegree{}) = \(\dfrac{FP}{9{,}7}\)

On a donc FP = 9,7 \(\times\) sin(60\textdegree{}) \(\approx\) 8,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VJP}\)) =  \(\dfrac{VP}{JP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VJP}\)) = \(\dfrac{4{,}8}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{VJP}\) =  ArcSin(\(\dfrac{4{,}8}{7{,}9}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.541) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BDF}\)) = \(\dfrac{BF}{FD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BDF}\)) =\(\dfrac{1{,}6}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BDF}\) = ArcSin(\(\dfrac{1{,}6}{8{,}8}\)) \(\approx\) 10\textdegree{}\end{document}
