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\begin{document}

\sloppy

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

 Dans le triangle NJF rectangle en N, on sait que :
\begin{itemize}
\item NJ=0,4 cm
\item \(\widehat{JFN}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle GRM rectangle en G, on sait que :
\begin{itemize}
\item GR=2,6 cm
\item RM=9,3 cm
\end{itemize}

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

 Dans le triangle VAB rectangle en V, on sait que :
\begin{itemize}
\item VA=3,1 cm
\item AB=6,9 cm
\end{itemize}

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

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

 Dans le triangle JDM rectangle en J, on sait que :
\begin{itemize}
\item JM=5,8 cm
\item \(\widehat{DMJ}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAZ rectangle en B, on sait que :
\begin{itemize}
\item BZ=8,7 cm
\item \(\widehat{BAZ}\) = 56\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [ZA]. (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 (5.293,1.157) {20°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NFJ}\)) = \(\dfrac{NJ}{NF}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{0{,}4}{NF}\)
\par\addvspace{.5em}
On a donc NF = \(\dfrac{0{,}4}{\tan(20^{\circ})}\) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 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(\(\widehat{GRM}\)) = \(\dfrac{2{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{GRM}\) =ArcCos(\(\dfrac{2{,}6}{9{,}3}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.301,0.855) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VBA}\)) = \(\dfrac{VA}{AB}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VBA}\)) =\(\dfrac{3{,}1}{6{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VBA}\) = ArcSin(\(\dfrac{3{,}1}{6{,}9}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.200) {19°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JMD}\)) = \(\dfrac{JD}{JM}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{JD}{5{,}8}\)
\par\addvspace{.5em}
On a donc JD =  5,8 \(\times\) tan(19\textdegree{})  \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {8,7 cm};
  \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 (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BAZ}\)) =  \(\dfrac{BZ}{AZ}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{8{,}7}{AZ}\)
\par\addvspace{.5em}
 On a donc AZ = \(\dfrac{8{,}7}{\sin(56^{\circ})}\) \(\approx\) 10,5 cm\end{document}
