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\begin{document}

\sloppy

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

 Dans le triangle FZD rectangle en F, on sait que :
\begin{itemize}
\item FZ=1,9 cm
\item FD=4,8 cm
\end{itemize}

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

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

 Dans le triangle ZNL rectangle en Z, on sait que :
\begin{itemize}
\item ZN=9,1 cm
\item \(\widehat{NLZ}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle BHS rectangle en B, on sait que :
\begin{itemize}
\item HS=0,9 cm
\item \(\widehat{HSB}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle KMF rectangle en K, on sait que :
\begin{itemize}
\item MF=0,6 cm
\item \(\widehat{KMF}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTZ rectangle en N, on sait que :
\begin{itemize}
\item NZ=4,7 cm
\item TZ=8,4 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FDZ}\)) = \(\dfrac{FZ}{FD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FDZ}\)) =  \(\dfrac{1{,}9}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FDZ}\) =  ArcTan(\(\dfrac{1{,}9}{4{,}8}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZLN}\)) = \(\dfrac{ZN}{NL}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{9{,}1}{NL}\)
\par\addvspace{.5em}
On a donc NL =\(\dfrac{9{,}1}{\sin(28^{\circ})}\) \(\approx\) 19,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BSH}\)) = \(\dfrac{BH}{HS}\)
\par\addvspace{.5em}
 D'où sin(30\textdegree{}) = \(\dfrac{BH}{0{,}9}\)

On a donc BH = 0,9 \(\times\) sin(30\textdegree{}) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{KMF}\)) =  \(\dfrac{KF}{MF}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{KF}{0{,}6}\)

On a donc KF = 0,6 \(\times\) sin(56\textdegree{}) \(\approx\) 0,5 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) {8,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.750,2.696) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NTZ}\)) =  \(\dfrac{NZ}{TZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NTZ}\)) = \(\dfrac{4{,}7}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{NTZ}\) =  ArcSin(\(\dfrac{4{,}7}{8{,}4}\)) \(\approx\) 34\textdegree{}\end{document}
