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\begin{document}

\sloppy

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

 Dans le triangle FKD rectangle en F, on sait que :
\begin{itemize}
\item FD=4,6 cm
\item KD=8,8 cm
\end{itemize}

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

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

 Dans le triangle WTB rectangle en W, on sait que :
\begin{itemize}
\item WB=1,9 cm
\item \(\widehat{TBW}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle GCB rectangle en G, on sait que :
\begin{itemize}
\item CB=7,4 cm
\item \(\widehat{CBG}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle NKC rectangle en N, on sait que :
\begin{itemize}
\item NC=4,9 cm
\item \(\widehat{KCN}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle FKS rectangle en F, on sait que :
\begin{itemize}
\item FK=2,8 cm
\item FS=6,6 cm
\end{itemize}

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.777,0.859) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FDK}\)) = \(\dfrac{FD}{KD}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{FDK}\)) =  \(\dfrac{4{,}6}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{FDK}\) = Arccos (\(\dfrac{4{,}6}{8{,}8}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WBT}\)) = \(\dfrac{WT}{WB}\)
\par\addvspace{.5em}
d'où  tan(29\textdegree{}) =  \(\dfrac{WT}{1{,}9}\)
\par\addvspace{.5em}
On a donc WT =  1,9 \(\times\) tan(29\textdegree{})  \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{GBC}\)) = \(\dfrac{GB}{CB}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{GB}{7{,}4}\)

On a donc GB = 7,4 \(\times\) cos(25\textdegree{}) \(\approx\) 6{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.017,0.723) {33°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{NCK}\)) = \(\dfrac{NC}{KC}\)
\par\addvspace{.5em}
  d'où    cos(33\textdegree{}) =  \(\dfrac{4{,}9}{KC}\)
\par\addvspace{.5em}
 On a donc KC =  \(\dfrac{4{,}9}{\cos(33^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.627) {6,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.418,2.417) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FKS}\)) = \(\dfrac{FS}{FK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{FKS}\)) = \(\dfrac{6{,}6}{2{,}8}\)
\par\addvspace{.5em}
 On a donc FKS = ArcTan(\(\dfrac{6{,}6}{2{,}8}\)) \(\approx\) 67\textdegree{}\end{document}
