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\begin{document}

\sloppy

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

 Dans le triangle ZWL rectangle en Z, on sait que :
\begin{itemize}
\item WL=7,7 cm
\item \(\widehat{WLZ}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle SFV rectangle en S, on sait que :
\begin{itemize}
\item SF=3,3 cm
\item FV=8,1 cm
\end{itemize}

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

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

 Dans le triangle JFH rectangle en J, on sait que :
\begin{itemize}
\item FH=4,4 cm
\item \(\widehat{FHJ}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle HNR rectangle en H, on sait que :
\begin{itemize}
\item HN=7,6 cm
\item \(\widehat{HNR}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJF rectangle en M, on sait que :
\begin{itemize}
\item MJ=2,7 cm
\item JF=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MJF}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZLW}\)) = \(\dfrac{ZW}{WL}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{ZW}{7{,}7}\)

On a donc ZW = 7,7 \(\times\) sin(23\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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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 l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{SVF}\)) = \(\dfrac{SF}{FV}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{SVF}\)) =\(\dfrac{3{,}3}{8{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SVF}\) = ArcSin(\(\dfrac{3{,}3}{8{,}1}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {?};
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\end{tikzpicture}\end{center}


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

cos(\(\widehat{JHF}\)) = \(\dfrac{JH}{FH}\)
\par\addvspace{.5em}
  d'où    cos(15\textdegree{}) =  \(\dfrac{JH}{4{,}4}\)

On a donc JH = 4,4 \(\times\) cos(15\textdegree{}) \(\approx\) 4{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HNR}\)) = \(\dfrac{HN}{NR}\)
\par\addvspace{.5em}
 d'où cos(73\textdegree{}) = \(\dfrac{7{,}6}{NR}\)
\par\addvspace{.5em}
On a donc NR = \(\dfrac{7{,}6}{\cos(73^{\circ})}\)\(\approx\) 26 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) {2,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.372) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MJF}\)) = \(\dfrac{MJ}{JF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MJF}\)) = \(\dfrac{2{,}7}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{MJF}\) =ArcCos(\(\dfrac{2{,}7}{7{,}2}\)) \(\approx\)68\textdegree{}\end{document}
