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\begin{document}

\sloppy

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

 Dans le triangle PRJ rectangle en P, on sait que :
\begin{itemize}
\item PR=2,5 cm
\item RJ=8,9 cm
\end{itemize}

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

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

 Dans le triangle CVZ rectangle en C, on sait que :
\begin{itemize}
\item CZ=9,3 cm
\item \(\widehat{VZC}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle SKL rectangle en S, on sait que :
\begin{itemize}
\item SL=9,9 cm
\item \(\widehat{SKL}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle LGH rectangle en L, on sait que :
\begin{itemize}
\item LG=7,7 cm
\item \(\widehat{LGH}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle HWZ rectangle en H, on sait que :
\begin{itemize}
\item HW=2,9 cm
\item WZ=7,2 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PJR}\)) = \(\dfrac{PR}{RJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PJR}\)) =\(\dfrac{2{,}5}{8{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PJR}\) = ArcSin(\(\dfrac{2{,}5}{8{,}9}\)) \(\approx\) 16\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CZV}\)) = \(\dfrac{CV}{CZ}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{CV}{9{,}3}\)
\par\addvspace{.5em}
On a donc CV =  9,3 \(\times\) tan(40\textdegree{})  \(\approx\) 7,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{SKL}\)) =  \(\dfrac{SL}{KL}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{9{,}9}{KL}\)
\par\addvspace{.5em}
 On a donc KL = \(\dfrac{9{,}9}{\sin(57^{\circ})}\) \(\approx\) 11,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LGH}\)) = \(\dfrac{LH}{LG}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{LH}{7{,}7}\)
\par\addvspace{.5em}
On a donc LH = 7,7 \(\times\) tan(77\textdegree{}) \(\approx\) 33,4 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) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.414,2.453) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HWZ}\)) = \(\dfrac{HW}{WZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HWZ}\)) = \(\dfrac{2{,}9}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{HWZ}\) =ArcCos(\(\dfrac{2{,}9}{7{,}2}\)) \(\approx\)66\textdegree{}\end{document}
