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\begin{document}

\sloppy

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

 Dans le triangle LWZ rectangle en L, on sait que :
\begin{itemize}
\item WZ=4,3 cm
\item \(\widehat{WZL}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle CKZ rectangle en C, on sait que :
\begin{itemize}
\item CK=1,2 cm
\item \(\widehat{KZC}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle NWT rectangle en N, on sait que :
\begin{itemize}
\item NT=2 cm
\item \(\widehat{NWT}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle CVK rectangle en C, on sait que :
\begin{itemize}
\item CV=2,4 cm
\item VK=8,7 cm
\end{itemize}

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

 Dans le triangle PWJ rectangle en P, on sait que :
\begin{itemize}
\item PW=2,1 cm
\item WJ=8,4 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LZW}\)) = \(\dfrac{LW}{WZ}\)
\par\addvspace{.5em}
 D'où sin(34\textdegree{}) = \(\dfrac{LW}{4{,}3}\)

On a donc LW = 4,3 \(\times\) sin(34\textdegree{}) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CZK}\)) = \(\dfrac{CK}{CZ}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{1{,}2}{CZ}\)
\par\addvspace{.5em}
On a donc CZ = \(\dfrac{1{,}2}{\tan(32^{\circ})}\) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{NWT}\)) =  \(\dfrac{NT}{WT}\)
\par\addvspace{.5em}
 d'où sin(65\textdegree{}) = \(\dfrac{2}{WT}\)
\par\addvspace{.5em}
 On a donc WT = \(\dfrac{2}{\sin(65^{\circ})}\) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{CVK}\)) = \(\dfrac{CV}{VK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CVK}\)) = \(\dfrac{2{,}4}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{CVK}\) =ArcCos(\(\dfrac{2{,}4}{8{,}7}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.385) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PJW}\)) = \(\dfrac{PW}{WJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PJW}\)) =\(\dfrac{2{,}1}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PJW}\) = ArcSin(\(\dfrac{2{,}1}{8{,}4}\)) \(\approx\) 14\textdegree{}\end{document}
