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\begin{document}

\sloppy

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

 Dans le triangle CKL rectangle en C, on sait que :
\begin{itemize}
\item KL=4,6 cm
\item \(\widehat{CKL}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle TZC rectangle en T, on sait que :
\begin{itemize}
\item TZ=2,4 cm
\item TC=4,2 cm
\end{itemize}

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

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

 Dans le triangle SWM rectangle en S, on sait que :
\begin{itemize}
\item SM=4,5 cm
\item \(\widehat{SWM}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLG rectangle en W, on sait que :
\begin{itemize}
\item WL=2,3 cm
\item LG=8,9 cm
\end{itemize}

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

 Dans le triangle WPT rectangle en W, on sait que :
\begin{itemize}
\item WT=1,4 cm
\item \(\widehat{PTW}\) =  23\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CKL}\)) =  \(\dfrac{CL}{KL}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{CL}{4{,}6}\)

On a donc CL = 4,6 \(\times\) sin(50\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TZC rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TCZ}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TCZ}\)) = \(\dfrac{TZ}{TC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TCZ}\)) =  \(\dfrac{2{,}4}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TCZ}\) =  ArcTan(\(\dfrac{2{,}4}{4{,}2}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{SWM}\)) = \(\dfrac{SM}{SW}\)
\par\addvspace{.5em}
tan(78\textdegree{}) = \(\dfrac{4{,}5}{SW}\)
\par\addvspace{.5em}
 On a donc SW = \(\dfrac{4{,}5}{\tan(78^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{WLG}\)) = \(\dfrac{WL}{LG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{WLG}\)) = \(\dfrac{2{,}3}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{WLG}\) =ArcCos(\(\dfrac{2{,}3}{8{,}9}\)) \(\approx\)75\textdegree{}
\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WTP}\)) = \(\dfrac{WT}{PT}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{1{,}4}{PT}\)
\par\addvspace{.5em}
 On a donc PT =  \(\dfrac{1{,}4}{\cos(23^{\circ})}\) \(\approx\) 1,5 cm\end{document}
