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\begin{document}

\sloppy

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

 Dans le triangle LRS rectangle en L, on sait que :
\begin{itemize}
\item LR=1 cm
\item \(\widehat{RSL}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle PWG rectangle en P, on sait que :
\begin{itemize}
\item PW=1,3 cm
\item PG=4,3 cm
\end{itemize}

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

 Dans le triangle KCP rectangle en K, on sait que :
\begin{itemize}
\item KP=8,7 cm
\item \(\widehat{KCP}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle LGF rectangle en L, on sait que :
\begin{itemize}
\item LG=6,6 cm
\item \(\widehat{LGF}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle HCJ rectangle en H, on sait que :
\begin{itemize}
\item HJ=5,4 cm
\item CJ=8,6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LSR}\)) = \(\dfrac{LR}{RS}\)
\par\addvspace{.5em}
 D'où sin(44\textdegree{}) = \(\dfrac{1}{RS}\)
\par\addvspace{.5em}
On a donc RS =\(\dfrac{1}{\sin(44^{\circ})}\) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PWG}\)) = \(\dfrac{PG}{PW}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PWG}\)) = \(\dfrac{4{,}3}{1{,}3}\)
\par\addvspace{.5em}
 On a donc PWG = ArcTan(\(\dfrac{4{,}3}{1{,}3}\)) \(\approx\) 73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{KCP}\)) = \(\dfrac{KP}{KC}\)
\par\addvspace{.5em}
tan(70\textdegree{}) = \(\dfrac{8{,}7}{KC}\)
\par\addvspace{.5em}
 On a donc KC = \(\dfrac{8{,}7}{\tan(70^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LGF}\)) = \(\dfrac{LF}{LG}\)
\par\addvspace{.5em}
d'où tan(57\textdegree{}) =\(\dfrac{LF}{6{,}6}\)
\par\addvspace{.5em}
On a donc LF = 6,6 \(\times\) tan(57\textdegree{}) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HJC}\)) = \(\dfrac{HJ}{CJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{HJC}\)) =  \(\dfrac{5{,}4}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{HJC}\) = Arccos (\(\dfrac{5{,}4}{8{,}6}\)) \(\approx\) 51\textdegree{} \end{document}
