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\begin{document}

\sloppy

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

 Dans le triangle FBW rectangle en F, on sait que :
\begin{itemize}
\item FB=2,4 cm
\item \(\widehat{FBW}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle KJD rectangle en K, on sait que :
\begin{itemize}
\item KD=7,1 cm
\item \(\widehat{KJD}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZHC rectangle en Z, on sait que :
\begin{itemize}
\item ZC=4,8 cm
\item HC=6,8 cm
\end{itemize}

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

 Dans le triangle FAZ rectangle en F, on sait que :
\begin{itemize}
\item FZ=0,6 cm
\item \(\widehat{AZF}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle PTD rectangle en P, on sait que :
\begin{itemize}
\item PT=3,3 cm
\item TD=9,6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FBW}\)) = \(\dfrac{FW}{FB}\)
\par\addvspace{.5em}
d'où tan(61\textdegree{}) =\(\dfrac{FW}{2{,}4}\)
\par\addvspace{.5em}
On a donc FW = 2,4 \(\times\) tan(61\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KJD}\)) =  \(\dfrac{KD}{JD}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{7{,}1}{JD}\)
\par\addvspace{.5em}
 On a donc JD = \(\dfrac{7{,}1}{\sin(57^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZHC}\)) =  \(\dfrac{ZC}{HC}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ZHC}\)) = \(\dfrac{4{,}8}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZHC}\) =  ArcSin(\(\dfrac{4{,}8}{6{,}8}\)) \(\approx\) 45\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FZA}\)) = \(\dfrac{FA}{FZ}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{FA}{0{,}6}\)
\par\addvspace{.5em}
On a donc FA =  0,6 \(\times\) tan(20\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) {9,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.153) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PDT}\)) = \(\dfrac{PT}{TD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PDT}\)) =\(\dfrac{3{,}3}{9{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PDT}\) = ArcSin(\(\dfrac{3{,}3}{9{,}6}\)) \(\approx\) 20\textdegree{}\end{document}
