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\begin{document}

\sloppy

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

 Dans le triangle RZA rectangle en R, on sait que :
\begin{itemize}
\item RZ=2,1 cm
\item RA=6,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RZA}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle LVT rectangle en L, on sait que :
\begin{itemize}
\item LV=2 cm
\item LT=6,5 cm
\end{itemize}

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

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

 Dans le triangle PLD rectangle en P, on sait que :
\begin{itemize}
\item PL=9,3 cm
\item \(\widehat{LDP}\) =  39\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 CLA rectangle en C, on sait que :
\begin{itemize}
\item CA=7,4 cm
\item \(\widehat{LAC}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle SZB rectangle en S, on sait que :
\begin{itemize}
\item SB=3,9 cm
\item \(\widehat{SZB}\) = 48\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RZA}\)) = \(\dfrac{RA}{RZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{RZA}\)) = \(\dfrac{6{,}2}{2{,}1}\)
\par\addvspace{.5em}
 On a donc RZA = ArcTan(\(\dfrac{6{,}2}{2{,}1}\)) \(\approx\) 71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LTV}\)) = \(\dfrac{LV}{LT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LTV}\)) =  \(\dfrac{2}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LTV}\) =  ArcTan(\(\dfrac{2}{6{,}5}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{PDL}\)) = \(\dfrac{PL}{PD}\)
\par\addvspace{.5em}
d'où  tan(39\textdegree{}) =  \(\dfrac{9{,}3}{PD}\)
\par\addvspace{.5em}
On a donc PD = \(\dfrac{9{,}3}{\tan(39^{\circ})}\) \(\approx\) 11,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{CAL}\)) = \(\dfrac{CA}{LA}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{7{,}4}{LA}\)
\par\addvspace{.5em}
 On a donc LA =  \(\dfrac{7{,}4}{\cos(37^{\circ})}\) \(\approx\) 9,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.216,2.724) {48°};
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 Dans le triangle SZB rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SZB}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{SZB}\)) = \(\dfrac{SB}{SZ}\)
\par\addvspace{.5em}
tan(48\textdegree{}) = \(\dfrac{3{,}9}{SZ}\)
\par\addvspace{.5em}
 On a donc SZ = \(\dfrac{3{,}9}{\tan(48^{\circ})}\) \(\approx\) 3,5 cm\end{document}
