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\begin{document}

\sloppy

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

 Dans le triangle RJA rectangle en R, on sait que :
\begin{itemize}
\item RJ=1,3 cm
\item \(\widehat{JAR}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle DLT rectangle en D, on sait que :
\begin{itemize}
\item DL=1,1 cm
\item LT=8,3 cm
\end{itemize}

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

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

 Dans le triangle NTR rectangle en N, on sait que :
\begin{itemize}
\item NT=2,7 cm
\item \(\widehat{NTR}\) = 59\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 ZPS rectangle en Z, on sait que :
\begin{itemize}
\item ZP=1,4 cm
\item PS=9,5 cm
\end{itemize}

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

 Dans le triangle AVT rectangle en A, on sait que :
\begin{itemize}
\item AT=1,7 cm
\item \(\widehat{AVT}\) = 72\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{RAJ}\)) = \(\dfrac{RJ}{JA}\)
\par\addvspace{.5em}
 D'où sin(24\textdegree{}) = \(\dfrac{1{,}3}{JA}\)
\par\addvspace{.5em}
On a donc JA =\(\dfrac{1{,}3}{\sin(24^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DTL}\)) = \(\dfrac{DL}{LT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DTL}\)) =\(\dfrac{1{,}1}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DTL}\) = ArcSin(\(\dfrac{1{,}1}{8{,}3}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NTR}\)) = \(\dfrac{NR}{NT}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{NR}{2{,}7}\)
\par\addvspace{.5em}
On a donc NR = 2,7 \(\times\) tan(59\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ZPS}\)) = \(\dfrac{ZP}{PS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{ZPS}\)) = \(\dfrac{1{,}4}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZPS}\) =ArcCos(\(\dfrac{1{,}4}{9{,}5}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.871) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.442,2.190) {72°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AVT}\)) = \(\dfrac{AT}{AV}\)
\par\addvspace{.5em}
tan(72\textdegree{}) = \(\dfrac{1{,}7}{AV}\)
\par\addvspace{.5em}
 On a donc AV = \(\dfrac{1{,}7}{\tan(72^{\circ})}\) \(\approx\) 0,6 cm\end{document}
