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\begin{document}

\sloppy

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

 Dans le triangle GWK rectangle en G, on sait que :
\begin{itemize}
\item GW=1,1 cm
\item GK=6,4 cm
\end{itemize}

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

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

 Dans le triangle ZBP rectangle en Z, on sait que :
\begin{itemize}
\item ZP=9,8 cm
\item \(\widehat{ZBP}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle LZP rectangle en L, on sait que :
\begin{itemize}
\item ZP=0,5 cm
\item \(\widehat{LZP}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle KLD rectangle en K, on sait que :
\begin{itemize}
\item KL=2,2 cm
\item \(\widehat{LDK}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAF rectangle en B, on sait que :
\begin{itemize}
\item BF=5 cm
\item AF=8,1 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{GKW}\)) = \(\dfrac{GW}{GK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{GKW}\)) =  \(\dfrac{1{,}1}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GKW}\) =  ArcTan(\(\dfrac{1{,}1}{6{,}4}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZBP}\)) =  \(\dfrac{ZP}{BP}\)
\par\addvspace{.5em}
 d'où sin(79\textdegree{}) = \(\dfrac{9{,}8}{BP}\)
\par\addvspace{.5em}
 On a donc BP = \(\dfrac{9{,}8}{\sin(79^{\circ})}\) \(\approx\) 10 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{LZP}\)) = \(\dfrac{LZ}{ZP}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{LZ}{0{,}5}\)

On a donc LZ = 0,5 \(\times\) cos(66\textdegree{}) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KDL}\)) = \(\dfrac{KL}{KD}\)
\par\addvspace{.5em}
d'où  tan(17\textdegree{}) =  \(\dfrac{2{,}2}{KD}\)
\par\addvspace{.5em}
On a donc KD = \(\dfrac{2{,}2}{\tan(17^{\circ})}\) \(\approx\) 7,2 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) {8,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.619,2.703) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BAF}\)) =  \(\dfrac{BF}{AF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BAF}\)) = \(\dfrac{5}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{BAF}\) =  ArcSin(\(\dfrac{5}{8{,}1}\)) \(\approx\) 38\textdegree{}\end{document}
