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\begin{document}

\sloppy

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

 Dans le triangle MDT rectangle en M, on sait que :
\begin{itemize}
\item DT=6 cm
\item \(\widehat{DTM}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle BMC rectangle en B, on sait que :
\begin{itemize}
\item BM=7,4 cm
\item \(\widehat{MCB}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle FPM rectangle en F, on sait que :
\begin{itemize}
\item FM=6,2 cm
\item PM=8,8 cm
\end{itemize}

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

 Dans le triangle ACM rectangle en A, on sait que :
\begin{itemize}
\item AM=0,5 cm
\item \(\widehat{ACM}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle KBR rectangle en K, on sait que :
\begin{itemize}
\item KR=6,6 cm
\item BR=7 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{MTD}\)) = \(\dfrac{MT}{DT}\)
\par\addvspace{.5em}
  d'où    cos(36\textdegree{}) =  \(\dfrac{MT}{6}\)

On a donc MT = 6 \(\times\) cos(36\textdegree{}) \(\approx\) 4{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{BCM}\)) = \(\dfrac{BM}{MC}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{7{,}4}{MC}\)
\par\addvspace{.5em}
On a donc MC =\(\dfrac{7{,}4}{\sin(23^{\circ})}\) \(\approx\) 18,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{FPM}\)) =  \(\dfrac{FM}{PM}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FPM}\)) = \(\dfrac{6{,}2}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{FPM}\) =  ArcSin(\(\dfrac{6{,}2}{8{,}8}\)) \(\approx\) 45\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{ACM}\)) = \(\dfrac{AM}{AC}\)
\par\addvspace{.5em}
tan(53\textdegree{}) = \(\dfrac{0{,}5}{AC}\)
\par\addvspace{.5em}
 On a donc AC = \(\dfrac{0{,}5}{\tan(53^{\circ})}\) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.180) {?};
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 Dans le triangle KBR rectangle en K, on cherche une relation entre l'angle aigu   \(\widehat{KRB}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{KRB}\)) = \(\dfrac{KR}{BR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KRB}\)) =  \(\dfrac{6{,}6}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{KRB}\) = Arccos (\(\dfrac{6{,}6}{7}\)) \(\approx\) 19\textdegree{} \end{document}
