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\begin{document}

\sloppy

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

 Dans le triangle PZD rectangle en P, on sait que :
\begin{itemize}
\item PD=7,4 cm
\item \(\widehat{ZDP}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle AKG rectangle en A, on sait que :
\begin{itemize}
\item KG=8,1 cm
\item \(\widehat{KGA}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle SFC rectangle en S, on sait que :
\begin{itemize}
\item SF=2,7 cm
\item SC=4,8 cm
\end{itemize}

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

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

 Dans le triangle PLJ rectangle en P, on sait que :
\begin{itemize}
\item LJ=6,8 cm
\item \(\widehat{PLJ}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle APB rectangle en A, on sait que :
\begin{itemize}
\item AP=2,8 cm
\item PB=10 cm
\end{itemize}

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

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

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

cos(\(\widehat{PDZ}\)) = \(\dfrac{PD}{ZD}\)
\par\addvspace{.5em}
  d'où    cos(31\textdegree{}) =  \(\dfrac{7{,}4}{ZD}\)
\par\addvspace{.5em}
 On a donc ZD =  \(\dfrac{7{,}4}{\cos(31^{\circ})}\) \(\approx\) 8,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{AGK}\)) = \(\dfrac{AG}{KG}\)
\par\addvspace{.5em}
  d'où    cos(16\textdegree{}) =  \(\dfrac{AG}{8{,}1}\)

On a donc AG = 8,1 \(\times\) cos(16\textdegree{}) \(\approx\) 7{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SCF}\)) = \(\dfrac{SF}{SC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SCF}\)) =  \(\dfrac{2{,}7}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SCF}\) =  ArcTan(\(\dfrac{2{,}7}{4{,}8}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{PLJ}\)) = \(\dfrac{PL}{LJ}\)
\par\addvspace{.5em}
d'où cos(49\textdegree{}) = \(\dfrac{PL}{6{,}8}\)

On a donc PL = 6,8 \(\times\) cos(49\textdegree{}) \(\approx\) 4,5 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) {10 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.450,2.115) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{APB}\)) = \(\dfrac{AP}{PB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{APB}\)) = \(\dfrac{2{,}8}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{APB}\) =ArcCos(\(\dfrac{2{,}8}{10}\)) \(\approx\)74\textdegree{}\end{document}
