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\begin{document}

\sloppy

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

 Dans le triangle GZD rectangle en G, on sait que :
\begin{itemize}
\item GD=3,9 cm
\item ZD=9,2 cm
\end{itemize}

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

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

 Dans le triangle TFS rectangle en T, on sait que :
\begin{itemize}
\item TS=8,7 cm
\item \(\widehat{TFS}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSD rectangle en K, on sait que :
\begin{itemize}
\item KS=0,6 cm
\item \(\widehat{SDK}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle RBS rectangle en R, on sait que :
\begin{itemize}
\item RB=9,9 cm
\item \(\widehat{BSR}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle FBK rectangle en F, on sait que :
\begin{itemize}
\item FK=4,5 cm
\item BK=9 cm
\end{itemize}

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

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

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

cos(\(\widehat{GDZ}\)) = \(\dfrac{GD}{ZD}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GDZ}\)) =  \(\dfrac{3{,}9}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{GDZ}\) = Arccos (\(\dfrac{3{,}9}{9{,}2}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TFS rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TFS}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TFS}\)) = \(\dfrac{TS}{TF}\)
\par\addvspace{.5em}
tan(55\textdegree{}) = \(\dfrac{8{,}7}{TF}\)
\par\addvspace{.5em}
 On a donc TF = \(\dfrac{8{,}7}{\tan(55^{\circ})}\) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{KDS}\)) = \(\dfrac{KS}{SD}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{0{,}6}{SD}\)
\par\addvspace{.5em}
On a donc SD =\(\dfrac{0{,}6}{\sin(21^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RSB}\)) = \(\dfrac{RB}{RS}\)
\par\addvspace{.5em}
d'où  tan(11\textdegree{}) =  \(\dfrac{9{,}9}{RS}\)
\par\addvspace{.5em}
On a donc RS = \(\dfrac{9{,}9}{\tan(11^{\circ})}\) \(\approx\) 50,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{FBK}\)) =  \(\dfrac{FK}{BK}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FBK}\)) = \(\dfrac{4{,}5}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{FBK}\) =  ArcSin(\(\dfrac{4{,}5}{9}\)) \(\approx\) 30\textdegree{}\end{document}
