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\begin{document}

\sloppy

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

 Dans le triangle BCD rectangle en B, on sait que :
\begin{itemize}
\item BD=5,1 cm
\item CD=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BCD}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle VZJ rectangle en V, on sait que :
\begin{itemize}
\item VJ=3,1 cm
\item \(\widehat{VZJ}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle SDA rectangle en S, on sait que :
\begin{itemize}
\item DA=2,8 cm
\item \(\widehat{SDA}\) = 46\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 KLJ rectangle en K, on sait que :
\begin{itemize}
\item KL=1,5 cm
\item KJ=6,4 cm
\end{itemize}

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

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

 Dans le triangle FZH rectangle en F, on sait que :
\begin{itemize}
\item FZ=1,5 cm
\item \(\widehat{ZHF}\) =  14\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [FH]. (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 (3.667,2.000) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.275,2.721) {?};
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 Dans le triangle BCD rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BCD}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{BCD}\)) =  \(\dfrac{BD}{CD}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BCD}\)) = \(\dfrac{5{,}1}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{BCD}\) =  ArcSin(\(\dfrac{5{,}1}{7}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VZJ}\)) =  \(\dfrac{VJ}{ZJ}\)
\par\addvspace{.5em}
 d'où sin(49\textdegree{}) = \(\dfrac{3{,}1}{ZJ}\)
\par\addvspace{.5em}
 On a donc ZJ = \(\dfrac{3{,}1}{\sin(49^{\circ})}\) \(\approx\) 4,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{SDA}\)) = \(\dfrac{SD}{DA}\)
\par\addvspace{.5em}
d'où cos(46\textdegree{}) = \(\dfrac{SD}{2{,}8}\)

On a donc SD = 2,8 \(\times\) cos(46\textdegree{}) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KJL}\)) = \(\dfrac{KL}{KJ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KJL}\)) =  \(\dfrac{1{,}5}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KJL}\) =  ArcTan(\(\dfrac{1{,}5}{6{,}4}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.056) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.404) {14°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FHZ}\)) = \(\dfrac{FZ}{FH}\)
\par\addvspace{.5em}
d'où  tan(14\textdegree{}) =  \(\dfrac{1{,}5}{FH}\)
\par\addvspace{.5em}
On a donc FH = \(\dfrac{1{,}5}{\tan(14^{\circ})}\) \(\approx\) 6 cm\end{document}
