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\begin{document}

\sloppy

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

 Dans le triangle LVB rectangle en L, on sait que :
\begin{itemize}
\item LV=1,6 cm
\item VB=9,3 cm
\end{itemize}

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

 Dans le triangle SHZ rectangle en S, on sait que :
\begin{itemize}
\item SZ=4,1 cm
\item \(\widehat{HZS}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle CMS rectangle en C, on sait que :
\begin{itemize}
\item CM=8,2 cm
\item \(\widehat{MSC}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle JRB rectangle en J, on sait que :
\begin{itemize}
\item JB=5,6 cm
\item RB=9,8 cm
\end{itemize}

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

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

 Dans le triangle HCV rectangle en H, on sait que :
\begin{itemize}
\item HC=4,5 cm
\item \(\widehat{CVH}\) =  14\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{LVB}\)) = \(\dfrac{LV}{VB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LVB}\)) = \(\dfrac{1{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{LVB}\) =ArcCos(\(\dfrac{1{,}6}{9{,}3}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{SZH}\)) = \(\dfrac{SH}{SZ}\)
\par\addvspace{.5em}
d'où  tan(12\textdegree{}) =  \(\dfrac{SH}{4{,}1}\)
\par\addvspace{.5em}
On a donc SH =  4,1 \(\times\) tan(12\textdegree{})  \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CSM}\)) = \(\dfrac{CM}{CS}\)
\par\addvspace{.5em}
d'où  tan(42\textdegree{}) =  \(\dfrac{8{,}2}{CS}\)
\par\addvspace{.5em}
On a donc CS = \(\dfrac{8{,}2}{\tan(42^{\circ})}\) \(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{JBR}\)) = \(\dfrac{JB}{RB}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JBR}\)) =  \(\dfrac{5{,}6}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{JBR}\) = Arccos (\(\dfrac{5{,}6}{9{,}8}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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


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

\par\addvspace{.5em}
sin(\(\widehat{HVC}\)) = \(\dfrac{HC}{CV}\)
\par\addvspace{.5em}
 D'où sin(14\textdegree{}) = \(\dfrac{4{,}5}{CV}\)
\par\addvspace{.5em}
On a donc CV =\(\dfrac{4{,}5}{\sin(14^{\circ})}\) \(\approx\) 18,6 cm\end{document}
