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\begin{document}

\sloppy

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

 Dans le triangle WJA rectangle en W, on sait que :
\begin{itemize}
\item WJ=1 cm
\item WA=6,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{WJA}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle NJK rectangle en N, on sait que :
\begin{itemize}
\item JK=6,1 cm
\item \(\widehat{NJK}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle JPT rectangle en J, on sait que :
\begin{itemize}
\item JT=9,3 cm
\item \(\widehat{PTJ}\) =  39\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFJ rectangle en D, on sait que :
\begin{itemize}
\item DF=3,3 cm
\item FJ=9,1 cm
\end{itemize}

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

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

 Dans le triangle DPG rectangle en D, on sait que :
\begin{itemize}
\item DG=3,2 cm
\item \(\widehat{DPG}\) = 52\textdegree{}
\end{itemize}

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

\newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{WJA}\)) = \(\dfrac{WA}{WJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WJA}\)) = \(\dfrac{6{,}5}{1}\)
\par\addvspace{.5em}
 On a donc WJA = ArcTan(6,5) \(\approx\) 81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NJK}\)) =  \(\dfrac{NK}{JK}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{NK}{6{,}1}\)

On a donc NK = 6,1 \(\times\) sin(67\textdegree{}) \(\approx\) 5,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JTP}\)) = \(\dfrac{JP}{JT}\)
\par\addvspace{.5em}
d'où  tan(39\textdegree{}) =  \(\dfrac{JP}{9{,}3}\)
\par\addvspace{.5em}
On a donc JP =  9,3 \(\times\) tan(39\textdegree{})  \(\approx\) 7,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DJF}\)) = \(\dfrac{DF}{FJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DJF}\)) =\(\dfrac{3{,}3}{9{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DJF}\) = ArcSin(\(\dfrac{3{,}3}{9{,}1}\)) \(\approx\) 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.000,2.734) {52°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{DPG}\)) =  \(\dfrac{DG}{PG}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{3{,}2}{PG}\)
\par\addvspace{.5em}
 On a donc PG = \(\dfrac{3{,}2}{\sin(52^{\circ})}\) \(\approx\) 4,1 cm\end{document}
