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\begin{document}

\sloppy

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

 Dans le triangle SVP rectangle en S, on sait que :
\begin{itemize}
\item SV=1,3 cm
\item SP=5,2 cm
\end{itemize}

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

 Dans le triangle SDR rectangle en S, on sait que :
\begin{itemize}
\item SD=7,6 cm
\item \(\widehat{DRS}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBD rectangle en C, on sait que :
\begin{itemize}
\item CB=3,1 cm
\item CD=3,9 cm
\end{itemize}

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

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

 Dans le triangle SFV rectangle en S, on sait que :
\begin{itemize}
\item SV=9,1 cm
\item \(\widehat{FVS}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle JSN rectangle en J, on sait que :
\begin{itemize}
\item JN=6,1 cm
\item \(\widehat{JSN}\) = 77\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [NS]. (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 (1.460,2.020) {?};
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 Dans le triangle SVP rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SVP}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{SVP}\)) = \(\dfrac{SP}{SV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{SVP}\)) = \(\dfrac{5{,}2}{1{,}3}\)
\par\addvspace{.5em}
 On a donc SVP = ArcTan(\(\dfrac{5{,}2}{1{,}3}\)) \(\approx\) 76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CDB}\)) = \(\dfrac{CB}{CD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CDB}\)) =  \(\dfrac{3{,}1}{3{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CDB}\) =  ArcTan(\(\dfrac{3{,}1}{3{,}9}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.101) {6,1 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 (1.465,1.977) {77°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JSN}\)) =  \(\dfrac{JN}{SN}\)
\par\addvspace{.5em}
 d'où sin(77\textdegree{}) = \(\dfrac{6{,}1}{SN}\)
\par\addvspace{.5em}
 On a donc SN = \(\dfrac{6{,}1}{\sin(77^{\circ})}\) \(\approx\) 6,3 cm\end{document}
