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\begin{document}

\sloppy

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

 Dans le triangle KGL rectangle en K, on sait que :
\begin{itemize}
\item KG=4,6 cm
\item \(\widehat{KGL}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle MVR rectangle en M, on sait que :
\begin{itemize}
\item MV=4,9 cm
\item \(\widehat{VRM}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle RKJ rectangle en R, on sait que :
\begin{itemize}
\item RK=2,6 cm
\item RJ=3,8 cm
\end{itemize}

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

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

 Dans le triangle NAG rectangle en N, on sait que :
\begin{itemize}
\item AG=1,2 cm
\item \(\widehat{NAG}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle GRM rectangle en G, on sait que :
\begin{itemize}
\item GR=1,8 cm
\item RM=8,9 cm
\end{itemize}

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

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

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

cos(\(\widehat{KGL}\)) = \(\dfrac{KG}{GL}\)
\par\addvspace{.5em}
 d'où cos(75\textdegree{}) = \(\dfrac{4{,}6}{GL}\)
\par\addvspace{.5em}
On a donc GL = \(\dfrac{4{,}6}{\cos(75^{\circ})}\)\(\approx\) 17,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MRV}\)) = \(\dfrac{MV}{MR}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{4{,}9}{MR}\)
\par\addvspace{.5em}
On a donc MR = \(\dfrac{4{,}9}{\tan(33^{\circ})}\) \(\approx\) 7,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RJK}\)) = \(\dfrac{RK}{RJ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RJK}\)) =  \(\dfrac{2{,}6}{3{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RJK}\) =  ArcTan(\(\dfrac{2{,}6}{3{,}8}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NAG}\)) = \(\dfrac{NA}{AG}\)
\par\addvspace{.5em}
d'où cos(64\textdegree{}) = \(\dfrac{NA}{1{,}2}\)

On a donc NA = 1,2 \(\times\) cos(64\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.471,1.922) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GRM}\)) = \(\dfrac{GR}{RM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GRM}\)) = \(\dfrac{1{,}8}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GRM}\) =ArcCos(\(\dfrac{1{,}8}{8{,}9}\)) \(\approx\)78\textdegree{}\end{document}
