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\begin{document}

\sloppy

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

 Dans le triangle KJR rectangle en K, on sait que :
\begin{itemize}
\item KJ=3,7 cm
\item \(\widehat{KJR}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle ACL rectangle en A, on sait que :
\begin{itemize}
\item CL=3 cm
\item \(\widehat{ACL}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle RJM rectangle en R, on sait que :
\begin{itemize}
\item RJ=5,1 cm
\item \(\widehat{RJM}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle BPM rectangle en B, on sait que :
\begin{itemize}
\item BM=4,1 cm
\item PM=8,1 cm
\end{itemize}

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

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

 Dans le triangle KMN rectangle en K, on sait que :
\begin{itemize}
\item KM=3,2 cm
\item MN=9,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KMN}\). \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 (1.737,2.745) {56°};
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 Dans le triangle KJR rectangle en K, on cherche une relation entre l'angle aigu   \(\widehat{KJR}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{KJR}\)) = \(\dfrac{KR}{KJ}\)
\par\addvspace{.5em}
d'où tan(56\textdegree{}) =\(\dfrac{KR}{3{,}7}\)
\par\addvspace{.5em}
On a donc KR = 3,7 \(\times\) tan(56\textdegree{}) \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ACL}\)) = \(\dfrac{AC}{CL}\)
\par\addvspace{.5em}
d'où cos(63\textdegree{}) = \(\dfrac{AC}{3}\)

On a donc AC = 3 \(\times\) cos(63\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{RJM}\)) = \(\dfrac{RJ}{JM}\)
\par\addvspace{.5em}
 d'où cos(74\textdegree{}) = \(\dfrac{5{,}1}{JM}\)
\par\addvspace{.5em}
On a donc JM = \(\dfrac{5{,}1}{\cos(74^{\circ})}\)\(\approx\) 18,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BMP}\)) = \(\dfrac{BM}{PM}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BMP}\)) =  \(\dfrac{4{,}1}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{BMP}\) = Arccos (\(\dfrac{4{,}1}{8{,}1}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.438,2.228) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KMN}\)) = \(\dfrac{KM}{MN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KMN}\)) = \(\dfrac{3{,}2}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{KMN}\) =ArcCos(\(\dfrac{3{,}2}{9{,}9}\)) \(\approx\)71\textdegree{}\end{document}
