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\begin{document}

\sloppy

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

 Dans le triangle NZT rectangle en N, on sait que :
\begin{itemize}
\item NT=6,6 cm
\item \(\widehat{ZTN}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle HWL rectangle en H, on sait que :
\begin{itemize}
\item HL=0,6 cm
\item \(\widehat{WLH}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle SPG rectangle en S, on sait que :
\begin{itemize}
\item SP=2,2 cm
\item \(\widehat{SPG}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle PMJ rectangle en P, on sait que :
\begin{itemize}
\item PJ=6,2 cm
\item MJ=7,3 cm
\end{itemize}

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

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

 Dans le triangle VGA rectangle en V, on sait que :
\begin{itemize}
\item VG=1,3 cm
\item GA=8,6 cm
\end{itemize}

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

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

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

cos(\(\widehat{NTZ}\)) = \(\dfrac{NT}{ZT}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{6{,}6}{ZT}\)
\par\addvspace{.5em}
 On a donc ZT =  \(\dfrac{6{,}6}{\cos(25^{\circ})}\) \(\approx\) 7,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HLW}\)) = \(\dfrac{HW}{HL}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{HW}{0{,}6}\)
\par\addvspace{.5em}
On a donc HW =  0,6 \(\times\) tan(32\textdegree{})  \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{SPG}\)) = \(\dfrac{SG}{SP}\)
\par\addvspace{.5em}
d'où tan(54\textdegree{}) =\(\dfrac{SG}{2{,}2}\)
\par\addvspace{.5em}
On a donc SG = 2,2 \(\times\) tan(54\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{PJM}\)) = \(\dfrac{PJ}{MJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{PJM}\)) =  \(\dfrac{6{,}2}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{PJM}\) = Arccos (\(\dfrac{6{,}2}{7{,}3}\)) \(\approx\) 32\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.484,1.802) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VGA}\)) = \(\dfrac{VG}{GA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VGA}\)) = \(\dfrac{1{,}3}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{VGA}\) =ArcCos(\(\dfrac{1{,}3}{8{,}6}\)) \(\approx\)81\textdegree{}\end{document}
