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\begin{document}

\sloppy

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

 Dans le triangle WBS rectangle en W, on sait que :
\begin{itemize}
\item WS=3,6 cm
\item BS=9,1 cm
\end{itemize}

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

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

 Dans le triangle DKS rectangle en D, on sait que :
\begin{itemize}
\item DK=2,1 cm
\item \(\widehat{DKS}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZGD rectangle en Z, on sait que :
\begin{itemize}
\item ZD=2,6 cm
\item \(\widehat{ZGD}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle KHZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=5,5 cm
\item HZ=8,3 cm
\end{itemize}

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

 Dans le triangle RKF rectangle en R, on sait que :
\begin{itemize}
\item KF=9 cm
\item \(\widehat{RKF}\) = 46\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{WSB}\)) = \(\dfrac{WS}{BS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{WSB}\)) =  \(\dfrac{3{,}6}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{WSB}\) = Arccos (\(\dfrac{3{,}6}{9{,}1}\)) \(\approx\) 67\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{DKS}\)) = \(\dfrac{DS}{DK}\)
\par\addvspace{.5em}
d'où tan(70\textdegree{}) =\(\dfrac{DS}{2{,}1}\)
\par\addvspace{.5em}
On a donc DS = 2,1 \(\times\) tan(70\textdegree{}) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ZGD}\)) =  \(\dfrac{ZD}{GD}\)
\par\addvspace{.5em}
 d'où sin(64\textdegree{}) = \(\dfrac{2{,}6}{GD}\)
\par\addvspace{.5em}
 On a donc GD = \(\dfrac{2{,}6}{\sin(64^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{KHZ}\)) =  \(\dfrac{KZ}{HZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KHZ}\)) = \(\dfrac{5{,}5}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{KHZ}\) =  ArcSin(\(\dfrac{5{,}5}{8{,}3}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{RKF}\)) = \(\dfrac{RK}{KF}\)
\par\addvspace{.5em}
d'où cos(46\textdegree{}) = \(\dfrac{RK}{9}\)

On a donc RK = 9 \(\times\) cos(46\textdegree{}) \(\approx\) 6,3 cm\end{document}
