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\begin{document}

\sloppy

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

 Dans le triangle FKG rectangle en F, on sait que :
\begin{itemize}
\item FG=8,2 cm
\item \(\widehat{KGF}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle DKR rectangle en D, on sait que :
\begin{itemize}
\item DR=6,6 cm
\item KR=8,2 cm
\end{itemize}

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

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

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

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

 Dans le triangle MNA rectangle en M, on sait que :
\begin{itemize}
\item NA=9,1 cm
\item \(\widehat{NAM}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle CTF rectangle en C, on sait que :
\begin{itemize}
\item CF=3,9 cm
\item \(\widehat{CTF}\) = 63\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FGK}\)) = \(\dfrac{FK}{FG}\)
\par\addvspace{.5em}
d'où  tan(35\textdegree{}) =  \(\dfrac{FK}{8{,}2}\)
\par\addvspace{.5em}
On a donc FK =  8,2 \(\times\) tan(35\textdegree{})  \(\approx\) 5,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DRK}\)) = \(\dfrac{DR}{KR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DRK}\)) =  \(\dfrac{6{,}6}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{DRK}\) = Arccos (\(\dfrac{6{,}6}{8{,}2}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VJF}\)) =  \(\dfrac{VF}{JF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VJF}\)) = \(\dfrac{3{,}9}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{VJF}\) =  ArcSin(\(\dfrac{3{,}9}{8{,}6}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MAN}\)) = \(\dfrac{MA}{NA}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{MA}{9{,}1}\)

On a donc MA = 9,1 \(\times\) cos(45\textdegree{}) \(\approx\) 6{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{CTF}\)) =  \(\dfrac{CF}{TF}\)
\par\addvspace{.5em}
 d'où sin(63\textdegree{}) = \(\dfrac{3{,}9}{TF}\)
\par\addvspace{.5em}
 On a donc TF = \(\dfrac{3{,}9}{\sin(63^{\circ})}\) \(\approx\) 4,4 cm\end{document}
