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\begin{document}

\sloppy

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

 Dans le triangle NFV rectangle en N, on sait que :
\begin{itemize}
\item NF=2,1 cm
\item NV=5,2 cm
\end{itemize}

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

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

 Dans le triangle WSD rectangle en W, on sait que :
\begin{itemize}
\item WD=5,6 cm
\item \(\widehat{SDW}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle KTR rectangle en K, on sait que :
\begin{itemize}
\item KR=6,4 cm
\item TR=8,8 cm
\end{itemize}

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

 Dans le triangle AFV rectangle en A, on sait que :
\begin{itemize}
\item AF=0,7 cm
\item \(\widehat{FVA}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle DJV rectangle en D, on sait que :
\begin{itemize}
\item DJ=9,1 cm
\item \(\widehat{JVD}\) =  15\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NVF}\)) = \(\dfrac{NF}{NV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NVF}\)) =  \(\dfrac{2{,}1}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NVF}\) =  ArcTan(\(\dfrac{2{,}1}{5{,}2}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WDS}\)) = \(\dfrac{WS}{WD}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{WS}{5{,}6}\)
\par\addvspace{.5em}
On a donc WS =  5,6 \(\times\) tan(32\textdegree{})  \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KTR}\)) =  \(\dfrac{KR}{TR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KTR}\)) = \(\dfrac{6{,}4}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KTR}\) =  ArcSin(\(\dfrac{6{,}4}{8{,}8}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{AVF}\)) = \(\dfrac{AF}{FV}\)
\par\addvspace{.5em}
 D'où sin(24\textdegree{}) = \(\dfrac{0{,}7}{FV}\)
\par\addvspace{.5em}
On a donc FV =\(\dfrac{0{,}7}{\sin(24^{\circ})}\) \(\approx\) 1,7 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) {9,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DVJ}\)) = \(\dfrac{DJ}{DV}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{9{,}1}{DV}\)
\par\addvspace{.5em}
On a donc DV = \(\dfrac{9{,}1}{\tan(15^{\circ})}\) \(\approx\) 34 cm\end{document}
