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\begin{document}

\sloppy

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

 Dans le triangle HTJ rectangle en H, on sait que :
\begin{itemize}
\item HT=3,1 cm
\item HJ=5,3 cm
\end{itemize}

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

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

 Dans le triangle HDK rectangle en H, on sait que :
\begin{itemize}
\item HD=7,1 cm
\item \(\widehat{HDK}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle VLA rectangle en V, on sait que :
\begin{itemize}
\item VA=5,2 cm
\item \(\widehat{VLA}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle SZF rectangle en S, on sait que :
\begin{itemize}
\item ZF=9,3 cm
\item \(\widehat{SZF}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFJ rectangle en T, on sait que :
\begin{itemize}
\item TJ=5,3 cm
\item FJ=8,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TFJ}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{HJT}\)) = \(\dfrac{HT}{HJ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HJT}\)) =  \(\dfrac{3{,}1}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HJT}\) =  ArcTan(\(\dfrac{3{,}1}{5{,}3}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HDK}\)) = \(\dfrac{HK}{HD}\)
\par\addvspace{.5em}
d'où tan(51\textdegree{}) =\(\dfrac{HK}{7{,}1}\)
\par\addvspace{.5em}
On a donc HK = 7,1 \(\times\) tan(51\textdegree{}) \(\approx\) 8,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VLA}\)) =  \(\dfrac{VA}{LA}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{5{,}2}{LA}\)
\par\addvspace{.5em}
 On a donc LA = \(\dfrac{5{,}2}{\sin(71^{\circ})}\) \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SZF}\)) = \(\dfrac{SZ}{ZF}\)
\par\addvspace{.5em}
d'où cos(75\textdegree{}) = \(\dfrac{SZ}{9{,}3}\)

On a donc SZ = 9,3 \(\times\) cos(75\textdegree{}) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{TFJ}\)) =  \(\dfrac{TJ}{FJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{TFJ}\)) = \(\dfrac{5{,}3}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{TFJ}\) =  ArcSin(\(\dfrac{5{,}3}{8{,}6}\)) \(\approx\) 38\textdegree{}\end{document}
