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\begin{document}

\sloppy

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

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

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

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

 Dans le triangle FZK rectangle en F, on sait que :
\begin{itemize}
\item FZ=3,3 cm
\item ZK=8,2 cm
\end{itemize}

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

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

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

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

 Dans le triangle GFP rectangle en G, on sait que :
\begin{itemize}
\item GF=6,7 cm
\item \(\widehat{GFP}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle PZD rectangle en P, on sait que :
\begin{itemize}
\item PZ=1,5 cm
\item ZD=7,6 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{LJD}\)) = \(\dfrac{LJ}{JD}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{LJ}{3{,}9}\)

On a donc LJ = 3,9 \(\times\) cos(56\textdegree{}) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{FZK}\)) = \(\dfrac{FZ}{ZK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{FZK}\)) = \(\dfrac{3{,}3}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{FZK}\) =ArcCos(\(\dfrac{3{,}3}{8{,}2}\)) \(\approx\)66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{PHD}\)) =  \(\dfrac{PD}{HD}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{PD}{6{,}8}\)

On a donc PD = 6,8 \(\times\) sin(73\textdegree{}) \(\approx\) 6,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{GFP}\)) = \(\dfrac{GF}{FP}\)
\par\addvspace{.5em}
 d'où cos(79\textdegree{}) = \(\dfrac{6{,}7}{FP}\)
\par\addvspace{.5em}
On a donc FP = \(\dfrac{6{,}7}{\cos(79^{\circ})}\)\(\approx\) 35,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.506) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PDZ}\)) = \(\dfrac{PZ}{ZD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PDZ}\)) =\(\dfrac{1{,}5}{7{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PDZ}\) = ArcSin(\(\dfrac{1{,}5}{7{,}6}\)) \(\approx\) 11\textdegree{}\end{document}
