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\begin{document}

\sloppy

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

 Dans le triangle BMP rectangle en B, on sait que :
\begin{itemize}
\item BP=5,1 cm
\item MP=6,8 cm
\end{itemize}

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

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

 Dans le triangle ZLB rectangle en Z, on sait que :
\begin{itemize}
\item LB=9,8 cm
\item \(\widehat{LBZ}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJV rectangle en F, on sait que :
\begin{itemize}
\item FJ=2,2 cm
\item \(\widehat{JVF}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle NBW rectangle en N, on sait que :
\begin{itemize}
\item NB=2,6 cm
\item NW=4,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NBW}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle AKF rectangle en A, on sait que :
\begin{itemize}
\item AF=7,7 cm
\item \(\widehat{KFA}\) =  35\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [AK]. (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 (4.464,0.769) {?};
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 Dans le triangle BMP rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BPM}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{BPM}\)) = \(\dfrac{BP}{MP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BPM}\)) =  \(\dfrac{5{,}1}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{BPM}\) = Arccos (\(\dfrac{5{,}1}{6{,}8}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZBL}\)) = \(\dfrac{ZB}{LB}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{ZB}{9{,}8}\)

On a donc ZB = 9,8 \(\times\) cos(40\textdegree{}) \(\approx\) 7{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{FVJ}\)) = \(\dfrac{FJ}{JV}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{2{,}2}{JV}\)
\par\addvspace{.5em}
On a donc JV =\(\dfrac{2{,}2}{\sin(31^{\circ})}\) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NBW}\)) = \(\dfrac{NW}{NB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{NBW}\)) = \(\dfrac{4{,}4}{2{,}6}\)
\par\addvspace{.5em}
 On a donc NBW = ArcTan(\(\dfrac{4{,}4}{2{,}6}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.864,0.734) {35°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AFK}\)) = \(\dfrac{AK}{AF}\)
\par\addvspace{.5em}
d'où  tan(35\textdegree{}) =  \(\dfrac{AK}{7{,}7}\)
\par\addvspace{.5em}
On a donc AK =  7,7 \(\times\) tan(35\textdegree{})  \(\approx\) 5,4 cm\end{document}
