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\begin{document}

\sloppy

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

 Dans le triangle CZT rectangle en C, on sait que :
\begin{itemize}
\item CZ=1,1 cm
\item ZT=8,9 cm
\end{itemize}

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

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

 Dans le triangle WBR rectangle en W, on sait que :
\begin{itemize}
\item WB=5,2 cm
\item \(\widehat{BRW}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAG rectangle en H, on sait que :
\begin{itemize}
\item AG=3,5 cm
\item \(\widehat{AGH}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle RAT rectangle en R, on sait que :
\begin{itemize}
\item RT=5,2 cm
\item AT=8,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RAT}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle VLP rectangle en V, on sait que :
\begin{itemize}
\item VL=0,9 cm
\item \(\widehat{VLP}\) = 79\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CTZ}\)) = \(\dfrac{CZ}{ZT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CTZ}\)) =\(\dfrac{1{,}1}{8{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CTZ}\) = ArcSin(\(\dfrac{1{,}1}{8{,}9}\)) \(\approx\) 7\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WRB}\)) = \(\dfrac{WB}{WR}\)
\par\addvspace{.5em}
d'où  tan(42\textdegree{}) =  \(\dfrac{5{,}2}{WR}\)
\par\addvspace{.5em}
On a donc WR = \(\dfrac{5{,}2}{\tan(42^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HGA}\)) = \(\dfrac{HA}{AG}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{HA}{3{,}5}\)

On a donc HA = 3,5 \(\times\) sin(19\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RAT}\)) =  \(\dfrac{RT}{AT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{RAT}\)) = \(\dfrac{5{,}2}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{RAT}\) =  ArcSin(\(\dfrac{5{,}2}{8{,}3}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.474,1.895) {79°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VLP}\)) = \(\dfrac{VL}{LP}\)
\par\addvspace{.5em}
 d'où cos(79\textdegree{}) = \(\dfrac{0{,}9}{LP}\)
\par\addvspace{.5em}
On a donc LP = \(\dfrac{0{,}9}{\cos(79^{\circ})}\)\(\approx\) 4,7 cm\end{document}
