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\begin{document}

\sloppy

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

 Dans le triangle ZKG rectangle en Z, on sait que :
\begin{itemize}
\item ZG=7,9 cm
\item \(\widehat{KGZ}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle AGF rectangle en A, on sait que :
\begin{itemize}
\item AG=3,7 cm
\item \(\widehat{AGF}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle DHS rectangle en D, on sait que :
\begin{itemize}
\item DS=6,5 cm
\item \(\widehat{DHS}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle AZH rectangle en A, on sait que :
\begin{itemize}
\item AZ=3,2 cm
\item ZH=8,8 cm
\end{itemize}

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

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

 Dans le triangle AWG rectangle en A, on sait que :
\begin{itemize}
\item AW=2,2 cm
\item WG=6,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AWG}\). \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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZGK}\)) = \(\dfrac{ZG}{KG}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{7{,}9}{KG}\)
\par\addvspace{.5em}
 On a donc KG =  \(\dfrac{7{,}9}{\cos(40^{\circ})}\) \(\approx\) 10,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{AGF}\)) = \(\dfrac{AF}{AG}\)
\par\addvspace{.5em}
d'où tan(67\textdegree{}) =\(\dfrac{AF}{3{,}7}\)
\par\addvspace{.5em}
On a donc AF = 3,7 \(\times\) tan(67\textdegree{}) \(\approx\) 8,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{DHS}\)) = \(\dfrac{DS}{DH}\)
\par\addvspace{.5em}
tan(65\textdegree{}) = \(\dfrac{6{,}5}{DH}\)
\par\addvspace{.5em}
 On a donc DH = \(\dfrac{6{,}5}{\tan(65^{\circ})}\) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{AHZ}\)) = \(\dfrac{AZ}{ZH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{AHZ}\)) =\(\dfrac{3{,}2}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AHZ}\) = ArcSin(\(\dfrac{3{,}2}{8{,}8}\)) \(\approx\) 21\textdegree{}
\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) {6,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.439,2.216) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{AWG}\)) = \(\dfrac{AW}{WG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{AWG}\)) = \(\dfrac{2{,}2}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{AWG}\) =ArcCos(\(\dfrac{2{,}2}{6{,}9}\)) \(\approx\)71\textdegree{}\end{document}
