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\begin{document}

\sloppy

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

 Dans le triangle FZS rectangle en F, on sait que :
\begin{itemize}
\item FS=5,9 cm
\item ZS=9,6 cm
\end{itemize}

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

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

 Dans le triangle WNH rectangle en W, on sait que :
\begin{itemize}
\item WH=0,8 cm
\item \(\widehat{WNH}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGP rectangle en T, on sait que :
\begin{itemize}
\item TG=1,9 cm
\item GP=9,7 cm
\end{itemize}

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

 Dans le triangle BNR rectangle en B, on sait que :
\begin{itemize}
\item NR=9,9 cm
\item \(\widehat{BNR}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle BCT rectangle en B, on sait que :
\begin{itemize}
\item BT=8,1 cm
\item \(\widehat{CTB}\) =  32\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [TC]. (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 (3.992,0.826) {?};
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 Dans le triangle FZS rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FSZ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{FSZ}\)) = \(\dfrac{FS}{ZS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{FSZ}\)) =  \(\dfrac{5{,}9}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{FSZ}\) = Arccos (\(\dfrac{5{,}9}{9{,}6}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WNH}\)) = \(\dfrac{WH}{WN}\)
\par\addvspace{.5em}
tan(78\textdegree{}) = \(\dfrac{0{,}8}{WN}\)
\par\addvspace{.5em}
 On a donc WN = \(\dfrac{0{,}8}{\tan(78^{\circ})}\) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{TGP}\)) = \(\dfrac{TG}{GP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TGP}\)) = \(\dfrac{1{,}9}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{TGP}\) =ArcCos(\(\dfrac{1{,}9}{9{,}7}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BNR}\)) =  \(\dfrac{BR}{NR}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{BR}{9{,}9}\)

On a donc BR = 9,9 \(\times\) sin(58\textdegree{}) \(\approx\) 8,4 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.100,0.717) {32°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BTC}\)) = \(\dfrac{BT}{CT}\)
\par\addvspace{.5em}
  d'où    cos(32\textdegree{}) =  \(\dfrac{8{,}1}{CT}\)
\par\addvspace{.5em}
 On a donc CT =  \(\dfrac{8{,}1}{\cos(32^{\circ})}\) \(\approx\) 9,6 cm\end{document}
