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\begin{document}

\sloppy

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

 Dans le triangle LCN rectangle en L, on sait que :
\begin{itemize}
\item LN=3,9 cm
\item \(\widehat{LCN}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle ABV rectangle en A, on sait que :
\begin{itemize}
\item AB=1,7 cm
\item BV=8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ABV}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle JAN rectangle en J, on sait que :
\begin{itemize}
\item AN=1,1 cm
\item \(\widehat{ANJ}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle JLC rectangle en J, on sait que :
\begin{itemize}
\item JL=2 cm
\item JC=6,5 cm
\end{itemize}

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

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

 Dans le triangle VPB rectangle en V, on sait que :
\begin{itemize}
\item PB=5,3 cm
\item \(\widehat{PBV}\) =  15\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [VP]. (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 (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LCN}\)) =  \(\dfrac{LN}{CN}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{3{,}9}{CN}\)
\par\addvspace{.5em}
 On a donc CN = \(\dfrac{3{,}9}{\sin(56^{\circ})}\) \(\approx\) 4,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ABV}\)) = \(\dfrac{AB}{BV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{ABV}\)) = \(\dfrac{1{,}7}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{ABV}\) =ArcCos(\(\dfrac{1{,}7}{8}\)) \(\approx\)78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{JNA}\)) = \(\dfrac{JN}{AN}\)
\par\addvspace{.5em}
  d'où    cos(14\textdegree{}) =  \(\dfrac{JN}{1{,}1}\)

On a donc JN = 1,1 \(\times\) cos(14\textdegree{}) \(\approx\) 1{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{JCL}\)) = \(\dfrac{JL}{JC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{JCL}\)) =  \(\dfrac{2}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JCL}\) =  ArcTan(\(\dfrac{2}{6{,}5}\)) \(\approx\) 17\textdegree{}
\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VBP}\)) = \(\dfrac{VP}{PB}\)
\par\addvspace{.5em}
 D'où sin(15\textdegree{}) = \(\dfrac{VP}{5{,}3}\)

On a donc VP = 5,3 \(\times\) sin(15\textdegree{}) \(\approx\) 1,4 cm\end{document}
