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\begin{document}

\sloppy

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

 Dans le triangle WDZ rectangle en W, on sait que :
\begin{itemize}
\item WZ=7,6 cm
\item \(\widehat{WDZ}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZFH rectangle en Z, on sait que :
\begin{itemize}
\item ZF=2,4 cm
\item FH=8,4 cm
\end{itemize}

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

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

 Dans le triangle HLS rectangle en H, on sait que :
\begin{itemize}
\item LS=8,1 cm
\item \(\widehat{HLS}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZHJ rectangle en Z, on sait que :
\begin{itemize}
\item ZJ=5,8 cm
\item HJ=9,5 cm
\end{itemize}

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

 Dans le triangle HMG rectangle en H, on sait que :
\begin{itemize}
\item MG=0,8 cm
\item \(\widehat{HMG}\) = 76\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{WDZ}\)) =  \(\dfrac{WZ}{DZ}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{7{,}6}{DZ}\)
\par\addvspace{.5em}
 On a donc DZ = \(\dfrac{7{,}6}{\sin(67^{\circ})}\) \(\approx\) 8,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZHF}\)) = \(\dfrac{ZF}{FH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZHF}\)) =\(\dfrac{2{,}4}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZHF}\) = ArcSin(\(\dfrac{2{,}4}{8{,}4}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{HLS}\)) = \(\dfrac{HL}{LS}\)
\par\addvspace{.5em}
d'où cos(68\textdegree{}) = \(\dfrac{HL}{8{,}1}\)

On a donc HL = 8,1 \(\times\) cos(68\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ZHJ}\)) =  \(\dfrac{ZJ}{HJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ZHJ}\)) = \(\dfrac{5{,}8}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZHJ}\) =  ArcSin(\(\dfrac{5{,}8}{9{,}5}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.460,2.019) {76°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HMG}\)) =  \(\dfrac{HG}{MG}\)
\par\addvspace{.5em}
 d'où sin(76\textdegree{}) = \(\dfrac{HG}{0{,}8}\)

On a donc HG = 0,8 \(\times\) sin(76\textdegree{}) \(\approx\) 0,8 cm\end{document}
