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\begin{document}

\sloppy

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

 Dans le triangle RHA rectangle en R, on sait que :
\begin{itemize}
\item RH=7,9 cm
\item \(\widehat{RHA}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle CAR rectangle en C, on sait que :
\begin{itemize}
\item CA=3 cm
\item \(\widehat{ARC}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle AWV rectangle en A, on sait que :
\begin{itemize}
\item AW=1,7 cm
\item WV=7,6 cm
\end{itemize}

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

 Dans le triangle PTH rectangle en P, on sait que :
\begin{itemize}
\item PT=1,4 cm
\item TH=7,7 cm
\end{itemize}

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

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

 Dans le triangle ZPV rectangle en Z, on sait que :
\begin{itemize}
\item ZV=8,8 cm
\item \(\widehat{PVZ}\) =  21\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RHA}\)) = \(\dfrac{RA}{RH}\)
\par\addvspace{.5em}
d'où tan(80\textdegree{}) =\(\dfrac{RA}{7{,}9}\)
\par\addvspace{.5em}
On a donc RA = 7,9 \(\times\) tan(80\textdegree{}) \(\approx\) 44,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{CRA}\)) = \(\dfrac{CA}{AR}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{3}{AR}\)
\par\addvspace{.5em}
On a donc AR =\(\dfrac{3}{\sin(41^{\circ})}\) \(\approx\) 4,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{AWV}\)) = \(\dfrac{AW}{WV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{AWV}\)) = \(\dfrac{1{,}7}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{AWV}\) =ArcCos(\(\dfrac{1{,}7}{7{,}6}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PHT}\)) = \(\dfrac{PT}{TH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PHT}\)) =\(\dfrac{1{,}4}{7{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PHT}\) = ArcSin(\(\dfrac{1{,}4}{7{,}7}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZVP}\)) = \(\dfrac{ZP}{ZV}\)
\par\addvspace{.5em}
d'où  tan(21\textdegree{}) =  \(\dfrac{ZP}{8{,}8}\)
\par\addvspace{.5em}
On a donc ZP =  8,8 \(\times\) tan(21\textdegree{})  \(\approx\) 3,4 cm\end{document}
