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\begin{document}

\sloppy

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

 Dans le triangle FLH rectangle en F, on sait que :
\begin{itemize}
\item FL=2,3 cm
\item LH=7,8 cm
\end{itemize}

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

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

 Dans le triangle LAG rectangle en L, on sait que :
\begin{itemize}
\item LA=2,6 cm
\item AG=7,1 cm
\end{itemize}

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

 Dans le triangle ZAB rectangle en Z, on sait que :
\begin{itemize}
\item AB=6,6 cm
\item \(\widehat{ABZ}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle GHJ rectangle en G, on sait que :
\begin{itemize}
\item HJ=3,4 cm
\item \(\widehat{GHJ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle PCF rectangle en P, on sait que :
\begin{itemize}
\item PC=8,6 cm
\item \(\widehat{PCF}\) = 73\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{FHL}\)) = \(\dfrac{FL}{LH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FHL}\)) =\(\dfrac{2{,}3}{7{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FHL}\) = ArcSin(\(\dfrac{2{,}3}{7{,}8}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{LAG}\)) = \(\dfrac{LA}{AG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LAG}\)) = \(\dfrac{2{,}6}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{LAG}\) =ArcCos(\(\dfrac{2{,}6}{7{,}1}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ZBA}\)) = \(\dfrac{ZA}{AB}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{ZA}{6{,}6}\)

On a donc ZA = 6,6 \(\times\) sin(28\textdegree{}) \(\approx\) 3,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GHJ}\)) =  \(\dfrac{GJ}{HJ}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{GJ}{3{,}4}\)

On a donc GJ = 3,4 \(\times\) sin(68\textdegree{}) \(\approx\) 3,2 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PCF}\)) = \(\dfrac{PC}{CF}\)
\par\addvspace{.5em}
 d'où cos(73\textdegree{}) = \(\dfrac{8{,}6}{CF}\)
\par\addvspace{.5em}
On a donc CF = \(\dfrac{8{,}6}{\cos(73^{\circ})}\)\(\approx\) 29,4 cm\end{document}
