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\begin{document}

\sloppy

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

 Dans le triangle AFZ rectangle en A, on sait que :
\begin{itemize}
\item AZ=4,5 cm
\item FZ=7,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AFZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle CMR rectangle en C, on sait que :
\begin{itemize}
\item MR=3,7 cm
\item \(\widehat{MRC}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle WRN rectangle en W, on sait que :
\begin{itemize}
\item WR=8 cm
\item \(\widehat{WRN}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle KHL rectangle en K, on sait que :
\begin{itemize}
\item KH=2,3 cm
\item HL=7,5 cm
\end{itemize}

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

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

 Dans le triangle TBR rectangle en T, on sait que :
\begin{itemize}
\item TR=2,9 cm
\item \(\widehat{BRT}\) =  18\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [TB]. (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 (2.696,2.699) {?};
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 Dans le triangle AFZ rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{AFZ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{AFZ}\)) =  \(\dfrac{AZ}{FZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{AFZ}\)) = \(\dfrac{4{,}5}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{AFZ}\) =  ArcSin(\(\dfrac{4{,}5}{7{,}7}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{CRM}\)) = \(\dfrac{CR}{MR}\)
\par\addvspace{.5em}
  d'où    cos(12\textdegree{}) =  \(\dfrac{CR}{3{,}7}\)

On a donc CR = 3,7 \(\times\) cos(12\textdegree{}) \(\approx\) 3{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{WRN}\)) = \(\dfrac{WR}{RN}\)
\par\addvspace{.5em}
 d'où cos(66\textdegree{}) = \(\dfrac{8}{RN}\)
\par\addvspace{.5em}
On a donc RN = \(\dfrac{8}{\cos(66^{\circ})}\)\(\approx\) 19,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{KLH}\)) = \(\dfrac{KH}{HL}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KLH}\)) =\(\dfrac{2{,}3}{7{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KLH}\) = ArcSin(\(\dfrac{2{,}3}{7{,}5}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.871) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.242) {18°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TRB}\)) = \(\dfrac{TB}{TR}\)
\par\addvspace{.5em}
d'où  tan(18\textdegree{}) =  \(\dfrac{TB}{2{,}9}\)
\par\addvspace{.5em}
On a donc TB =  2,9 \(\times\) tan(18\textdegree{})  \(\approx\) 0,9 cm\end{document}
