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\begin{document}

\sloppy

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

 Dans le triangle CWD rectangle en C, on sait que :
\begin{itemize}
\item CD=6,5 cm
\item \(\widehat{CWD}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle BKZ rectangle en B, on sait que :
\begin{itemize}
\item BZ=6 cm
\item KZ=8,7 cm
\end{itemize}

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

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

 Dans le triangle HFB rectangle en H, on sait que :
\begin{itemize}
\item HF=8 cm
\item \(\widehat{FBH}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle SVC rectangle en S, on sait que :
\begin{itemize}
\item SC=6 cm
\item VC=7,7 cm
\end{itemize}

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

 Dans le triangle BDS rectangle en B, on sait que :
\begin{itemize}
\item DS=9,9 cm
\item \(\widehat{BDS}\) = 54\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CWD}\)) =  \(\dfrac{CD}{WD}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{6{,}5}{WD}\)
\par\addvspace{.5em}
 On a donc WD = \(\dfrac{6{,}5}{\sin(71^{\circ})}\) \(\approx\) 6,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{BZK}\)) = \(\dfrac{BZ}{KZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BZK}\)) =  \(\dfrac{6}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{BZK}\) = Arccos (\(\dfrac{6}{8{,}7}\)) \(\approx\) 46\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{HBF}\)) = \(\dfrac{HF}{HB}\)
\par\addvspace{.5em}
d'où  tan(44\textdegree{}) =  \(\dfrac{8}{HB}\)
\par\addvspace{.5em}
On a donc HB = \(\dfrac{8}{\tan(44^{\circ})}\) \(\approx\) 8,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{SVC}\)) =  \(\dfrac{SC}{VC}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{SVC}\)) = \(\dfrac{6}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SVC}\) =  ArcSin(\(\dfrac{6}{7{,}7}\)) \(\approx\) 51\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BDS}\)) = \(\dfrac{BD}{DS}\)
\par\addvspace{.5em}
d'où cos(54\textdegree{}) = \(\dfrac{BD}{9{,}9}\)

On a donc BD = 9,9 \(\times\) cos(54\textdegree{}) \(\approx\) 5,8 cm\end{document}
