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\begin{document}

\sloppy

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

 Dans le triangle SLR rectangle en S, on sait que :
\begin{itemize}
\item SR=2,7 cm
\item \(\widehat{SLR}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle CKR rectangle en C, on sait que :
\begin{itemize}
\item CR=5,1 cm
\item KR=10 cm
\end{itemize}

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

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

 Dans le triangle KLZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=4,8 cm
\item \(\widehat{LZK}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle RTP rectangle en R, on sait que :
\begin{itemize}
\item RT=10 cm
\item \(\widehat{RTP}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle LAJ rectangle en L, on sait que :
\begin{itemize}
\item LJ=5,7 cm
\item AJ=8 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SLR}\)) =  \(\dfrac{SR}{LR}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{2{,}7}{LR}\)
\par\addvspace{.5em}
 On a donc LR = \(\dfrac{2{,}7}{\sin(56^{\circ})}\) \(\approx\) 3,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{CRK}\)) = \(\dfrac{CR}{KR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{CRK}\)) =  \(\dfrac{5{,}1}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{CRK}\) = Arccos (\(\dfrac{5{,}1}{10}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{KZL}\)) = \(\dfrac{KL}{KZ}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{KL}{4{,}8}\)
\par\addvspace{.5em}
On a donc KL =  4,8 \(\times\) tan(40\textdegree{})  \(\approx\) 4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RTP}\)) = \(\dfrac{RP}{RT}\)
\par\addvspace{.5em}
d'où tan(48\textdegree{}) =\(\dfrac{RP}{10}\)
\par\addvspace{.5em}
On a donc RP = 10 \(\times\) tan(48\textdegree{}) \(\approx\) 11,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{LAJ}\)) =  \(\dfrac{LJ}{AJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LAJ}\)) = \(\dfrac{5{,}7}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{LAJ}\) =  ArcSin(\(\dfrac{5{,}7}{8}\)) \(\approx\) 45\textdegree{}\end{document}
