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\begin{document}

\sloppy

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

 Dans le triangle ZGW rectangle en Z, on sait que :
\begin{itemize}
\item ZW=6,6 cm
\item GW=8,5 cm
\end{itemize}

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

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

 Dans le triangle VKH rectangle en V, on sait que :
\begin{itemize}
\item VK=2,3 cm
\item \(\widehat{KHV}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle TBG rectangle en T, on sait que :
\begin{itemize}
\item TG=1,5 cm
\item \(\widehat{BGT}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle PHG rectangle en P, on sait que :
\begin{itemize}
\item HG=8,1 cm
\item \(\widehat{PHG}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle KAN rectangle en K, on sait que :
\begin{itemize}
\item KA=3,1 cm
\item AN=8,3 cm
\end{itemize}

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

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

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

cos(\(\widehat{ZWG}\)) = \(\dfrac{ZW}{GW}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZWG}\)) =  \(\dfrac{6{,}6}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZWG}\) = Arccos (\(\dfrac{6{,}6}{8{,}5}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VHK}\)) = \(\dfrac{VK}{KH}\)
\par\addvspace{.5em}
 D'où sin(20\textdegree{}) = \(\dfrac{2{,}3}{KH}\)
\par\addvspace{.5em}
On a donc KH =\(\dfrac{2{,}3}{\sin(20^{\circ})}\) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TGB}\)) = \(\dfrac{TB}{TG}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{TB}{1{,}5}\)
\par\addvspace{.5em}
On a donc TB =  1,5 \(\times\) tan(33\textdegree{})  \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PHG}\)) =  \(\dfrac{PG}{HG}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{PG}{8{,}1}\)

On a donc PG = 8,1 \(\times\) sin(57\textdegree{}) \(\approx\) 6,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.367) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KAN}\)) = \(\dfrac{KA}{AN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KAN}\)) = \(\dfrac{3{,}1}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{KAN}\) =ArcCos(\(\dfrac{3{,}1}{8{,}3}\)) \(\approx\)68\textdegree{}\end{document}
