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\begin{document}

\sloppy

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

 Dans le triangle FTK rectangle en F, on sait que :
\begin{itemize}
\item FK=1,9 cm
\item \(\widehat{FTK}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle WRK rectangle en W, on sait que :
\begin{itemize}
\item WR=2,4 cm
\item WK=4,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{WRK}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle WCZ rectangle en W, on sait que :
\begin{itemize}
\item WC=2 cm
\item CZ=6,7 cm
\end{itemize}

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

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

 Dans le triangle FCS rectangle en F, on sait que :
\begin{itemize}
\item CS=5,4 cm
\item \(\widehat{FCS}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle KAZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=1,9 cm
\item \(\widehat{KAZ}\) = 46\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.371) {68°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FTK}\)) = \(\dfrac{FK}{FT}\)
\par\addvspace{.5em}
tan(68\textdegree{}) = \(\dfrac{1{,}9}{FT}\)
\par\addvspace{.5em}
 On a donc FT = \(\dfrac{1{,}9}{\tan(68^{\circ})}\) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WRK}\)) = \(\dfrac{WK}{WR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WRK}\)) = \(\dfrac{4{,}1}{2{,}4}\)
\par\addvspace{.5em}
 On a donc WRK = ArcTan(\(\dfrac{4{,}1}{2{,}4}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WZC}\)) = \(\dfrac{WC}{CZ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WZC}\)) =\(\dfrac{2}{6{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WZC}\) = ArcSin(\(\dfrac{2}{6{,}7}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{FCS}\)) =  \(\dfrac{FS}{CS}\)
\par\addvspace{.5em}
 d'où sin(72\textdegree{}) = \(\dfrac{FS}{5{,}4}\)

On a donc FS = 5,4 \(\times\) sin(72\textdegree{}) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {1,9 cm};
  \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 (2.311,2.720) {46°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KAZ}\)) =  \(\dfrac{KZ}{AZ}\)
\par\addvspace{.5em}
 d'où sin(46\textdegree{}) = \(\dfrac{1{,}9}{AZ}\)
\par\addvspace{.5em}
 On a donc AZ = \(\dfrac{1{,}9}{\sin(46^{\circ})}\) \(\approx\) 2,6 cm\end{document}
