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\begin{document}

\sloppy

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

 Dans le triangle PNM rectangle en P, on sait que :
\begin{itemize}
\item PN=2,7 cm
\item PM=5,7 cm
\end{itemize}

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

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

 Dans le triangle DMV rectangle en D, on sait que :
\begin{itemize}
\item DM=1 cm
\item MV=9,9 cm
\end{itemize}

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

 Dans le triangle LFD rectangle en L, on sait que :
\begin{itemize}
\item LF=3,5 cm
\item \(\widehat{FDL}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle VRJ rectangle en V, on sait que :
\begin{itemize}
\item VJ=4,1 cm
\item \(\widehat{VRJ}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle AGB rectangle en A, on sait que :
\begin{itemize}
\item GB=4,2 cm
\item \(\widehat{AGB}\) = 49\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{PMN}\)) = \(\dfrac{PN}{PM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{PMN}\)) =  \(\dfrac{2{,}7}{5{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PMN}\) =  ArcTan(\(\dfrac{2{,}7}{5{,}7}\)) \(\approx\) 25\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DMV}\)) = \(\dfrac{DM}{MV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DMV}\)) = \(\dfrac{1}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{DMV}\) =ArcCos(\(\dfrac{1}{9{,}9}\)) \(\approx\)84\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LDF}\)) = \(\dfrac{LF}{FD}\)
\par\addvspace{.5em}
 D'où sin(25\textdegree{}) = \(\dfrac{3{,}5}{FD}\)
\par\addvspace{.5em}
On a donc FD =\(\dfrac{3{,}5}{\sin(25^{\circ})}\) \(\approx\) 8,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VRJ}\)) = \(\dfrac{VJ}{VR}\)
\par\addvspace{.5em}
tan(53\textdegree{}) = \(\dfrac{4{,}1}{VR}\)
\par\addvspace{.5em}
 On a donc VR = \(\dfrac{4{,}1}{\tan(53^{\circ})}\) \(\approx\) 3,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.166,2.727) {49°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AGB}\)) =  \(\dfrac{AB}{GB}\)
\par\addvspace{.5em}
 d'où sin(49\textdegree{}) = \(\dfrac{AB}{4{,}2}\)

On a donc AB = 4,2 \(\times\) sin(49\textdegree{}) \(\approx\) 3,2 cm\end{document}
