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\begin{document}

\sloppy

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

 Dans le triangle JVD rectangle en J, on sait que :
\begin{itemize}
\item VD=1,5 cm
\item \(\widehat{VDJ}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle FKB rectangle en F, on sait que :
\begin{itemize}
\item FK=2,6 cm
\item FB=4,3 cm
\end{itemize}

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

 Dans le triangle SDF rectangle en S, on sait que :
\begin{itemize}
\item DF=4,2 cm
\item \(\widehat{SDF}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle BJD rectangle en B, on sait que :
\begin{itemize}
\item BD=1,4 cm
\item \(\widehat{JDB}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZJV rectangle en Z, on sait que :
\begin{itemize}
\item ZJ=2 cm
\item JV=9,2 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{JDV}\)) = \(\dfrac{JD}{VD}\)
\par\addvspace{.5em}
  d'où    cos(18\textdegree{}) =  \(\dfrac{JD}{1{,}5}\)

On a donc JD = 1,5 \(\times\) cos(18\textdegree{}) \(\approx\) 1{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{FKB}\)) = \(\dfrac{FB}{FK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{FKB}\)) = \(\dfrac{4{,}3}{2{,}6}\)
\par\addvspace{.5em}
 On a donc FKB = ArcTan(\(\dfrac{4{,}3}{2{,}6}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{SDF}\)) = \(\dfrac{SD}{DF}\)
\par\addvspace{.5em}
d'où cos(69\textdegree{}) = \(\dfrac{SD}{4{,}2}\)

On a donc SD = 4,2 \(\times\) cos(69\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BDJ}\)) = \(\dfrac{BD}{JD}\)
\par\addvspace{.5em}
  d'où    cos(33\textdegree{}) =  \(\dfrac{1{,}4}{JD}\)
\par\addvspace{.5em}
 On a donc JD =  \(\dfrac{1{,}4}{\cos(33^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.461) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZVJ}\)) = \(\dfrac{ZJ}{JV}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZVJ}\)) =\(\dfrac{2}{9{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZVJ}\) = ArcSin(\(\dfrac{2}{9{,}2}\)) \(\approx\) 13\textdegree{}\end{document}
