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\begin{document}

\sloppy

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

 Dans le triangle TKN rectangle en T, on sait que :
\begin{itemize}
\item KN=2 cm
\item \(\widehat{TKN}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle RCV rectangle en R, on sait que :
\begin{itemize}
\item RC=1,1 cm
\item \(\widehat{RCV}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle TZK rectangle en T, on sait que :
\begin{itemize}
\item TZ=2,8 cm
\item TK=4,6 cm
\end{itemize}

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

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

 Dans le triangle TCS rectangle en T, on sait que :
\begin{itemize}
\item CS=0,5 cm
\item \(\widehat{CST}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle BWL rectangle en B, on sait que :
\begin{itemize}
\item BL=4,9 cm
\item WL=8,1 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{TKN}\)) =  \(\dfrac{TN}{KN}\)
\par\addvspace{.5em}
 d'où sin(74\textdegree{}) = \(\dfrac{TN}{2}\)

On a donc TN = 2 \(\times\) sin(74\textdegree{}) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RCV}\)) = \(\dfrac{RC}{CV}\)
\par\addvspace{.5em}
 d'où cos(75\textdegree{}) = \(\dfrac{1{,}1}{CV}\)
\par\addvspace{.5em}
On a donc CV = \(\dfrac{1{,}1}{\cos(75^{\circ})}\)\(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TKZ}\)) = \(\dfrac{TZ}{TK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TKZ}\)) =  \(\dfrac{2{,}8}{4{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TKZ}\) =  ArcTan(\(\dfrac{2{,}8}{4{,}6}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle TCS rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TSC}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TSC}\)) = \(\dfrac{TC}{CS}\)
\par\addvspace{.5em}
 D'où sin(27\textdegree{}) = \(\dfrac{TC}{0{,}5}\)

On a donc TC = 0,5 \(\times\) sin(27\textdegree{}) \(\approx\) 0,2 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) {8,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.649,2.702) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BWL}\)) =  \(\dfrac{BL}{WL}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BWL}\)) = \(\dfrac{4{,}9}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{BWL}\) =  ArcSin(\(\dfrac{4{,}9}{8{,}1}\)) \(\approx\) 37\textdegree{}\end{document}
