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\begin{document}

\sloppy

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

 Dans le triangle CZR rectangle en C, on sait que :
\begin{itemize}
\item CZ=4,3 cm
\item \(\widehat{CZR}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle AHB rectangle en A, on sait que :
\begin{itemize}
\item AH=7,3 cm
\item \(\widehat{HBA}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle ADN rectangle en A, on sait que :
\begin{itemize}
\item AN=3,5 cm
\item \(\widehat{ADN}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle LSP rectangle en L, on sait que :
\begin{itemize}
\item LS=2,8 cm
\item LP=3,5 cm
\end{itemize}

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

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

 Dans le triangle PAT rectangle en P, on sait que :
\begin{itemize}
\item PA=2,7 cm
\item AT=8 cm
\end{itemize}

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

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

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

cos(\(\widehat{CZR}\)) = \(\dfrac{CZ}{ZR}\)
\par\addvspace{.5em}
 d'où cos(46\textdegree{}) = \(\dfrac{4{,}3}{ZR}\)
\par\addvspace{.5em}
On a donc ZR = \(\dfrac{4{,}3}{\cos(46^{\circ})}\)\(\approx\) 6,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ABH}\)) = \(\dfrac{AH}{AB}\)
\par\addvspace{.5em}
d'où  tan(14\textdegree{}) =  \(\dfrac{7{,}3}{AB}\)
\par\addvspace{.5em}
On a donc AB = \(\dfrac{7{,}3}{\tan(14^{\circ})}\) \(\approx\) 29,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{ADN}\)) = \(\dfrac{AN}{AD}\)
\par\addvspace{.5em}
tan(80\textdegree{}) = \(\dfrac{3{,}5}{AD}\)
\par\addvspace{.5em}
 On a donc AD = \(\dfrac{3{,}5}{\tan(80^{\circ})}\) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LPS}\)) = \(\dfrac{LS}{LP}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LPS}\)) =  \(\dfrac{2{,}8}{3{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LPS}\) =  ArcTan(\(\dfrac{2{,}8}{3{,}5}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.434,2.267) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PAT}\)) = \(\dfrac{PA}{AT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PAT}\)) = \(\dfrac{2{,}7}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{PAT}\) =ArcCos(\(\dfrac{2{,}7}{8}\)) \(\approx\)70\textdegree{}\end{document}
