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\begin{document}

\sloppy

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

 Dans le triangle LFP rectangle en L, on sait que :
\begin{itemize}
\item LP=8,4 cm
\item \(\widehat{FPL}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle DSW rectangle en D, on sait que :
\begin{itemize}
\item SW=1 cm
\item \(\widehat{DSW}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle SFZ rectangle en S, on sait que :
\begin{itemize}
\item SF=1,2 cm
\item SZ=6,2 cm
\end{itemize}

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

 Dans le triangle JSV rectangle en J, on sait que :
\begin{itemize}
\item JS=3,7 cm
\item \(\widehat{JSV}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle GND rectangle en G, on sait que :
\begin{itemize}
\item GN=1,5 cm
\item ND=8,7 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LPF}\)) = \(\dfrac{LF}{LP}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{LF}{8{,}4}\)
\par\addvspace{.5em}
On a donc LF =  8,4 \(\times\) tan(23\textdegree{})  \(\approx\) 3,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DSW}\)) =  \(\dfrac{DW}{SW}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{DW}{1}\)

On a donc DW = 1 \(\times\) sin(58\textdegree{}) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SFZ}\)) = \(\dfrac{SZ}{SF}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{SFZ}\)) = \(\dfrac{6{,}2}{1{,}2}\)
\par\addvspace{.5em}
 On a donc SFZ = ArcTan(\(\dfrac{6{,}2}{1{,}2}\)) \(\approx\) 79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{JSV}\)) = \(\dfrac{JS}{SV}\)
\par\addvspace{.5em}
 d'où cos(52\textdegree{}) = \(\dfrac{3{,}7}{SV}\)
\par\addvspace{.5em}
On a donc SV = \(\dfrac{3{,}7}{\cos(52^{\circ})}\)\(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.562) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GDN}\)) = \(\dfrac{GN}{ND}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GDN}\)) =\(\dfrac{1{,}5}{8{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GDN}\) = ArcSin(\(\dfrac{1{,}5}{8{,}7}\)) \(\approx\) 10\textdegree{}\end{document}
