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\begin{document}

\sloppy

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

 Dans le triangle FPT rectangle en F, on sait que :
\begin{itemize}
\item FP=6,7 cm
\item \(\widehat{FPT}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle LHB rectangle en L, on sait que :
\begin{itemize}
\item LB=5,5 cm
\item HB=8,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LHB}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle CAF rectangle en C, on sait que :
\begin{itemize}
\item CF=0,4 cm
\item \(\widehat{CAF}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle BZK rectangle en B, on sait que :
\begin{itemize}
\item BK=4,4 cm
\item ZK=9,2 cm
\end{itemize}

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

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

 Dans le triangle JVK rectangle en J, on sait que :
\begin{itemize}
\item VK=7,9 cm
\item \(\widehat{VKJ}\) =  27\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FPT}\)) = \(\dfrac{FT}{FP}\)
\par\addvspace{.5em}
d'où tan(48\textdegree{}) =\(\dfrac{FT}{6{,}7}\)
\par\addvspace{.5em}
On a donc FT = 6,7 \(\times\) tan(48\textdegree{}) \(\approx\) 7,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{LHB}\)) =  \(\dfrac{LB}{HB}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LHB}\)) = \(\dfrac{5{,}5}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{LHB}\) =  ArcSin(\(\dfrac{5{,}5}{8{,}5}\)) \(\approx\) 40\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{CAF}\)) =  \(\dfrac{CF}{AF}\)
\par\addvspace{.5em}
 d'où sin(54\textdegree{}) = \(\dfrac{0{,}4}{AF}\)
\par\addvspace{.5em}
 On a donc AF = \(\dfrac{0{,}4}{\sin(54^{\circ})}\) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BKZ}\)) = \(\dfrac{BK}{ZK}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BKZ}\)) =  \(\dfrac{4{,}4}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{BKZ}\) = Arccos (\(\dfrac{4{,}4}{9{,}2}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.302,0.841) {27°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JKV}\)) = \(\dfrac{JV}{VK}\)
\par\addvspace{.5em}
 D'où sin(27\textdegree{}) = \(\dfrac{JV}{7{,}9}\)

On a donc JV = 7,9 \(\times\) sin(27\textdegree{}) \(\approx\) 3,6 cm\end{document}
