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\begin{document}

\sloppy

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

 Dans le triangle JVK rectangle en J, on sait que :
\begin{itemize}
\item VK=3 cm
\item \(\widehat{JVK}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle CFJ rectangle en C, on sait que :
\begin{itemize}
\item CF=4,7 cm
\item \(\widehat{FJC}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle TCF rectangle en T, on sait que :
\begin{itemize}
\item TC=3,1 cm
\item CF=7,3 cm
\end{itemize}

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

 Dans le triangle DRT rectangle en D, on sait que :
\begin{itemize}
\item DT=3,9 cm
\item \(\widehat{RTD}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle GZR rectangle en G, on sait que :
\begin{itemize}
\item GR=5,2 cm
\item ZR=10 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{JVK}\)) =  \(\dfrac{JK}{VK}\)
\par\addvspace{.5em}
 d'où sin(80\textdegree{}) = \(\dfrac{JK}{3}\)

On a donc JK = 3 \(\times\) sin(80\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{CJF}\)) = \(\dfrac{CF}{FJ}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{4{,}7}{FJ}\)
\par\addvspace{.5em}
On a donc FJ =\(\dfrac{4{,}7}{\sin(29^{\circ})}\) \(\approx\) 9,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.520) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{TCF}\)) = \(\dfrac{TC}{CF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TCF}\)) = \(\dfrac{3{,}1}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{TCF}\) =ArcCos(\(\dfrac{3{,}1}{7{,}3}\)) \(\approx\)65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DTR}\)) = \(\dfrac{DR}{DT}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{DR}{3{,}9}\)
\par\addvspace{.5em}
On a donc DR =  3,9 \(\times\) tan(19\textdegree{})  \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {10 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.771,0.860) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GRZ}\)) = \(\dfrac{GR}{ZR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GRZ}\)) =  \(\dfrac{5{,}2}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{GRZ}\) = Arccos (\(\dfrac{5{,}2}{10}\)) \(\approx\) 59\textdegree{} \end{document}
