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\begin{document}

\sloppy

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

 Dans le triangle LPT rectangle en L, on sait que :
\begin{itemize}
\item LP=3,1 cm
\item PT=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LPT}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle NRK rectangle en N, on sait que :
\begin{itemize}
\item NR=0,5 cm
\item \(\widehat{RKN}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle VBR rectangle en V, on sait que :
\begin{itemize}
\item BR=3,3 cm
\item \(\widehat{BRV}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle AFP rectangle en A, on sait que :
\begin{itemize}
\item AF=7,2 cm
\item \(\widehat{FPA}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle SKF rectangle en S, on sait que :
\begin{itemize}
\item SK=1 cm
\item KF=9 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{LPT}\)) = \(\dfrac{LP}{PT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LPT}\)) = \(\dfrac{3{,}1}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{LPT}\) =ArcCos(\(\dfrac{3{,}1}{7{,}2}\)) \(\approx\)64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NKR}\)) = \(\dfrac{NR}{NK}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{0{,}5}{NK}\)
\par\addvspace{.5em}
On a donc NK = \(\dfrac{0{,}5}{\tan(16^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VRB}\)) = \(\dfrac{VB}{BR}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{VB}{3{,}3}\)

On a donc VB = 3,3 \(\times\) sin(29\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{APF}\)) = \(\dfrac{AF}{FP}\)
\par\addvspace{.5em}
 D'où sin(37\textdegree{}) = \(\dfrac{7{,}2}{FP}\)
\par\addvspace{.5em}
On a donc FP =\(\dfrac{7{,}2}{\sin(37^{\circ})}\) \(\approx\) 12 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.284,1.697) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SFK}\)) = \(\dfrac{SK}{KF}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{SFK}\)) =\(\dfrac{1}{9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SFK}\) = ArcSin(\(\dfrac{1}{9}\)) \(\approx\) 6\textdegree{}\end{document}
