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\begin{document}

\sloppy

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

 Dans le triangle RVK rectangle en R, on sait que :
\begin{itemize}
\item RV=2,4 cm
\item VK=7,8 cm
\end{itemize}

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

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

 Dans le triangle KJR rectangle en K, on sait que :
\begin{itemize}
\item KJ=3,4 cm
\item \(\widehat{JRK}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle KJZ rectangle en K, on sait que :
\begin{itemize}
\item JZ=8,8 cm
\item \(\widehat{KJZ}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle BJW rectangle en B, on sait que :
\begin{itemize}
\item BW=9,2 cm
\item \(\widehat{BJW}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAJ rectangle en D, on sait que :
\begin{itemize}
\item DJ=5,1 cm
\item AJ=7 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{RKV}\)) = \(\dfrac{RV}{VK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RKV}\)) =\(\dfrac{2{,}4}{7{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RKV}\) = ArcSin(\(\dfrac{2{,}4}{7{,}8}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KRJ}\)) = \(\dfrac{KJ}{KR}\)
\par\addvspace{.5em}
d'où  tan(28\textdegree{}) =  \(\dfrac{3{,}4}{KR}\)
\par\addvspace{.5em}
On a donc KR = \(\dfrac{3{,}4}{\tan(28^{\circ})}\) \(\approx\) 6,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KJZ}\)) = \(\dfrac{KJ}{JZ}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{KJ}{8{,}8}\)

On a donc KJ = 8,8 \(\times\) cos(66\textdegree{}) \(\approx\) 3,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BJW}\)) =  \(\dfrac{BW}{JW}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{9{,}2}{JW}\)
\par\addvspace{.5em}
 On a donc JW = \(\dfrac{9{,}2}{\sin(51^{\circ})}\) \(\approx\) 11,8 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) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.275,2.721) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{DAJ}\)) =  \(\dfrac{DJ}{AJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DAJ}\)) = \(\dfrac{5{,}1}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{DAJ}\) =  ArcSin(\(\dfrac{5{,}1}{7}\)) \(\approx\) 47\textdegree{}\end{document}
