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\begin{document}

\sloppy

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

 Dans le triangle ALZ rectangle en A, on sait que :
\begin{itemize}
\item AL=6,3 cm
\item \(\widehat{LZA}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle JZD rectangle en J, on sait que :
\begin{itemize}
\item JZ=1 cm
\item JD=5,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{JZD}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MJT rectangle en M, on sait que :
\begin{itemize}
\item MT=6,3 cm
\item JT=7,1 cm
\end{itemize}

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

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

 Dans le triangle RWV rectangle en R, on sait que :
\begin{itemize}
\item RV=0,9 cm
\item \(\widehat{RWV}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFN rectangle en T, on sait que :
\begin{itemize}
\item TN=4,7 cm
\item \(\widehat{TFN}\) = 74\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{AZL}\)) = \(\dfrac{AL}{AZ}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{6{,}3}{AZ}\)
\par\addvspace{.5em}
On a donc AZ = \(\dfrac{6{,}3}{\tan(22^{\circ})}\) \(\approx\) 15,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{JZD}\)) = \(\dfrac{JD}{JZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JZD}\)) = \(\dfrac{5{,}4}{1}\)
\par\addvspace{.5em}
 On a donc JZD = ArcTan(5,4) \(\approx\) 80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{MTJ}\)) = \(\dfrac{MT}{JT}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MTJ}\)) =  \(\dfrac{6{,}3}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{MTJ}\) = Arccos (\(\dfrac{6{,}3}{7{,}1}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RWV}\)) =  \(\dfrac{RV}{WV}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{0{,}9}{WV}\)
\par\addvspace{.5em}
 On a donc WV = \(\dfrac{0{,}9}{\sin(67^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.451,2.103) {74°};
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 Dans le triangle TFN rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TFN}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TFN}\)) = \(\dfrac{TN}{TF}\)
\par\addvspace{.5em}
tan(74\textdegree{}) = \(\dfrac{4{,}7}{TF}\)
\par\addvspace{.5em}
 On a donc TF = \(\dfrac{4{,}7}{\tan(74^{\circ})}\) \(\approx\) 1,3 cm\end{document}
