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\begin{document}

\sloppy

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

 Dans le triangle FMN rectangle en F, on sait que :
\begin{itemize}
\item MN=3,1 cm
\item \(\widehat{FMN}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle PLJ rectangle en P, on sait que :
\begin{itemize}
\item PL=2,7 cm
\item PJ=4,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PLJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle THG rectangle en T, on sait que :
\begin{itemize}
\item TH=8,6 cm
\item \(\widehat{THG}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle DZN rectangle en D, on sait que :
\begin{itemize}
\item DZ=2,1 cm
\item ZN=9,5 cm
\end{itemize}

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

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

 Dans le triangle WJG rectangle en W, on sait que :
\begin{itemize}
\item JG=8,1 cm
\item \(\widehat{JGW}\) =  22\textdegree{}
\end{itemize}

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

\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 (1.498,2.753) {59°};
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 Dans le triangle FMN rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FMN}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{FMN}\)) = \(\dfrac{FM}{MN}\)
\par\addvspace{.5em}
d'où cos(59\textdegree{}) = \(\dfrac{FM}{3{,}1}\)

On a donc FM = 3,1 \(\times\) cos(59\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PLJ}\)) = \(\dfrac{PJ}{PL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PLJ}\)) = \(\dfrac{4{,}2}{2{,}7}\)
\par\addvspace{.5em}
 On a donc PLJ = ArcTan(\(\dfrac{4{,}2}{2{,}7}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{THG}\)) = \(\dfrac{TH}{HG}\)
\par\addvspace{.5em}
 d'où cos(53\textdegree{}) = \(\dfrac{8{,}6}{HG}\)
\par\addvspace{.5em}
On a donc HG = \(\dfrac{8{,}6}{\cos(53^{\circ})}\)\(\approx\) 14,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DNZ}\)) = \(\dfrac{DZ}{ZN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DNZ}\)) =\(\dfrac{2{,}1}{9{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DNZ}\) = ArcSin(\(\dfrac{2{,}1}{9{,}5}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WGJ}\)) = \(\dfrac{WG}{JG}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{WG}{8{,}1}\)

On a donc WG = 8,1 \(\times\) cos(22\textdegree{}) \(\approx\) 7{,}5 cm\end{document}
