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\begin{document}

\sloppy

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

 Dans le triangle RBC rectangle en R, on sait que :
\begin{itemize}
\item RB=2,9 cm
\item RC=5,5 cm
\end{itemize}

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

 Dans le triangle VNW rectangle en V, on sait que :
\begin{itemize}
\item VN=7,7 cm
\item \(\widehat{NWV}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZSB rectangle en Z, on sait que :
\begin{itemize}
\item ZS=1,4 cm
\item ZB=3,8 cm
\end{itemize}

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

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

 Dans le triangle HND rectangle en H, on sait que :
\begin{itemize}
\item ND=6 cm
\item \(\widehat{HND}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle BVA rectangle en B, on sait que :
\begin{itemize}
\item BV=8,2 cm
\item \(\widehat{VAB}\) =  12\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RBC}\)) = \(\dfrac{RC}{RB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{RBC}\)) = \(\dfrac{5{,}5}{2{,}9}\)
\par\addvspace{.5em}
 On a donc RBC = ArcTan(\(\dfrac{5{,}5}{2{,}9}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{VWN}\)) = \(\dfrac{VN}{VW}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{7{,}7}{VW}\)
\par\addvspace{.5em}
On a donc VW = \(\dfrac{7{,}7}{\tan(36^{\circ})}\) \(\approx\) 10,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZBS}\)) = \(\dfrac{ZS}{ZB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZBS}\)) =  \(\dfrac{1{,}4}{3{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZBS}\) =  ArcTan(\(\dfrac{1{,}4}{3{,}8}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HND}\)) = \(\dfrac{HN}{ND}\)
\par\addvspace{.5em}
d'où cos(75\textdegree{}) = \(\dfrac{HN}{6}\)

On a donc HN = 6 \(\times\) cos(75\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {8,2 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 (5.287,1.482) {12°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BAV}\)) = \(\dfrac{BV}{VA}\)
\par\addvspace{.5em}
 D'où sin(12\textdegree{}) = \(\dfrac{8{,}2}{VA}\)
\par\addvspace{.5em}
On a donc VA =\(\dfrac{8{,}2}{\sin(12^{\circ})}\) \(\approx\) 39,4 cm\end{document}
