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\begin{document}

\sloppy

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

 Dans le triangle VCA rectangle en V, on sait que :
\begin{itemize}
\item VA=6,3 cm
\item CA=8,9 cm
\end{itemize}

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

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

 Dans le triangle GPK rectangle en G, on sait que :
\begin{itemize}
\item GP=3,3 cm
\item \(\widehat{GPK}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle LRD rectangle en L, on sait que :
\begin{itemize}
\item RD=7,5 cm
\item \(\widehat{LRD}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle TMF rectangle en T, on sait que :
\begin{itemize}
\item TM=2,3 cm
\item TF=6,6 cm
\end{itemize}

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

 Dans le triangle WFD rectangle en W, on sait que :
\begin{itemize}
\item WD=4,5 cm
\item \(\widehat{WFD}\) = 58\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [DF]. (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 (4.287,0.788) {?};
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 Dans le triangle VCA rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VAC}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{VAC}\)) = \(\dfrac{VA}{CA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VAC}\)) =  \(\dfrac{6{,}3}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{VAC}\) = Arccos (\(\dfrac{6{,}3}{8{,}9}\)) \(\approx\) 45\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{GPK}\)) = \(\dfrac{GK}{GP}\)
\par\addvspace{.5em}
d'où tan(70\textdegree{}) =\(\dfrac{GK}{3{,}3}\)
\par\addvspace{.5em}
On a donc GK = 3,3 \(\times\) tan(70\textdegree{}) \(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{LRD}\)) = \(\dfrac{LR}{RD}\)
\par\addvspace{.5em}
d'où cos(72\textdegree{}) = \(\dfrac{LR}{7{,}5}\)

On a donc LR = 7,5 \(\times\) cos(72\textdegree{}) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TMF}\)) = \(\dfrac{TF}{TM}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{TMF}\)) = \(\dfrac{6{,}6}{2{,}3}\)
\par\addvspace{.5em}
 On a donc TMF = ArcTan(\(\dfrac{6{,}6}{2{,}3}\)) \(\approx\) 71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WFD}\)) =  \(\dfrac{WD}{FD}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{4{,}5}{FD}\)
\par\addvspace{.5em}
 On a donc FD = \(\dfrac{4{,}5}{\sin(58^{\circ})}\) \(\approx\) 5,3 cm\end{document}
