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\begin{document}

\sloppy

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

 Dans le triangle FMG rectangle en F, on sait que :
\begin{itemize}
\item FG=7,3 cm
\item \(\widehat{FMG}\) = 72\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 NKW rectangle en N, on sait que :
\begin{itemize}
\item NW=5,9 cm
\item KW=8 cm
\end{itemize}

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

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

 Dans le triangle JHK rectangle en J, on sait que :
\begin{itemize}
\item JK=1,6 cm
\item \(\widehat{HKJ}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle FRK rectangle en F, on sait que :
\begin{itemize}
\item RK=4,8 cm
\item \(\widehat{RKF}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle AJR rectangle en A, on sait que :
\begin{itemize}
\item AR=5,2 cm
\item JR=8,8 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FMG}\)) = \(\dfrac{FG}{FM}\)
\par\addvspace{.5em}
tan(72\textdegree{}) = \(\dfrac{7{,}3}{FM}\)
\par\addvspace{.5em}
 On a donc FM = \(\dfrac{7{,}3}{\tan(72^{\circ})}\) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{NWK}\)) = \(\dfrac{NW}{KW}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{NWK}\)) =  \(\dfrac{5{,}9}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{NWK}\) = Arccos (\(\dfrac{5{,}9}{8}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{JKH}\)) = \(\dfrac{JK}{HK}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{1{,}6}{HK}\)
\par\addvspace{.5em}
 On a donc HK =  \(\dfrac{1{,}6}{\cos(25^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{FKR}\)) = \(\dfrac{FK}{RK}\)
\par\addvspace{.5em}
  d'où    cos(18\textdegree{}) =  \(\dfrac{FK}{4{,}8}\)

On a donc FK = 4,8 \(\times\) cos(18\textdegree{}) \(\approx\) 4{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.682,2.700) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AJR}\)) =  \(\dfrac{AR}{JR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{AJR}\)) = \(\dfrac{5{,}2}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{AJR}\) =  ArcSin(\(\dfrac{5{,}2}{8{,}8}\)) \(\approx\) 36\textdegree{}\end{document}
