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\begin{document}

\sloppy

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

 Dans le triangle VAW rectangle en V, on sait que :
\begin{itemize}
\item VA=8,7 cm
\item \(\widehat{VAW}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle MKP rectangle en M, on sait que :
\begin{itemize}
\item MK=2,8 cm
\item MP=6,4 cm
\end{itemize}

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

 Dans le triangle PFD rectangle en P, on sait que :
\begin{itemize}
\item PD=3,9 cm
\item \(\widehat{FDP}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle HDW rectangle en H, on sait que :
\begin{itemize}
\item DW=3,8 cm
\item \(\widehat{HDW}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJS rectangle en F, on sait que :
\begin{itemize}
\item FS=5,4 cm
\item JS=8,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{VAW}\)) = \(\dfrac{VW}{VA}\)
\par\addvspace{.5em}
d'où tan(65\textdegree{}) =\(\dfrac{VW}{8{,}7}\)
\par\addvspace{.5em}
On a donc VW = 8,7 \(\times\) tan(65\textdegree{}) \(\approx\) 18,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MKP}\)) = \(\dfrac{MP}{MK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MKP}\)) = \(\dfrac{6{,}4}{2{,}8}\)
\par\addvspace{.5em}
 On a donc MKP = ArcTan(\(\dfrac{6{,}4}{2{,}8}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{PDF}\)) = \(\dfrac{PD}{FD}\)
\par\addvspace{.5em}
  d'où    cos(11\textdegree{}) =  \(\dfrac{3{,}9}{FD}\)
\par\addvspace{.5em}
 On a donc FD =  \(\dfrac{3{,}9}{\cos(11^{\circ})}\) \(\approx\) 4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HDW}\)) = \(\dfrac{HD}{DW}\)
\par\addvspace{.5em}
d'où cos(67\textdegree{}) = \(\dfrac{HD}{3{,}8}\)

On a donc HD = 3,8 \(\times\) cos(67\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.049,0.818) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FSJ}\)) = \(\dfrac{FS}{JS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{FSJ}\)) =  \(\dfrac{5{,}4}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{FSJ}\) = Arccos (\(\dfrac{5{,}4}{8{,}5}\)) \(\approx\) 51\textdegree{} \end{document}
