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\begin{document}

\sloppy

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

 Dans le triangle JAP rectangle en J, on sait que :
\begin{itemize}
\item AP=8,7 cm
\item \(\widehat{APJ}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle PLD rectangle en P, on sait que :
\begin{itemize}
\item PD=4,4 cm
\item \(\widehat{PLD}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle LRF rectangle en L, on sait que :
\begin{itemize}
\item LR=2 cm
\item LF=4,9 cm
\end{itemize}

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

 Dans le triangle BVH rectangle en B, on sait que :
\begin{itemize}
\item BV=2,3 cm
\item VH=6,8 cm
\end{itemize}

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

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

 Dans le triangle GTJ rectangle en G, on sait que :
\begin{itemize}
\item GT=0,8 cm
\item \(\widehat{TJG}\) =  31\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [JT]. (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 (4.726,0.745) {37°};
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 Dans le triangle JAP rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JPA}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{JPA}\)) = \(\dfrac{JP}{AP}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{JP}{8{,}7}\)

On a donc JP = 8,7 \(\times\) cos(37\textdegree{}) \(\approx\) 6{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PLD}\)) = \(\dfrac{PD}{PL}\)
\par\addvspace{.5em}
tan(66\textdegree{}) = \(\dfrac{4{,}4}{PL}\)
\par\addvspace{.5em}
 On a donc PL = \(\dfrac{4{,}4}{\tan(66^{\circ})}\) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{LRF}\)) = \(\dfrac{LF}{LR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{LRF}\)) = \(\dfrac{4{,}9}{2}\)
\par\addvspace{.5em}
 On a donc LRF = ArcTan(\(\dfrac{4{,}9}{2}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BHV}\)) = \(\dfrac{BV}{VH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BHV}\)) =\(\dfrac{2{,}3}{6{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BHV}\) = ArcSin(\(\dfrac{2{,}3}{6{,}8}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.188,0.711) {31°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GJT}\)) = \(\dfrac{GT}{TJ}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{0{,}8}{TJ}\)
\par\addvspace{.5em}
On a donc TJ =\(\dfrac{0{,}8}{\sin(31^{\circ})}\) \(\approx\) 1,6 cm\end{document}
