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\begin{document}

\sloppy

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

 Dans le triangle TNM rectangle en T, on sait que :
\begin{itemize}
\item TN=3,4 cm
\item \(\widehat{NMT}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle LFT rectangle en L, on sait que :
\begin{itemize}
\item LT=5,7 cm
\item FT=10 cm
\end{itemize}

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

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

 Dans le triangle AGP rectangle en A, on sait que :
\begin{itemize}
\item GP=1,4 cm
\item \(\widehat{GPA}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle MCH rectangle en M, on sait que :
\begin{itemize}
\item MC=2,5 cm
\item CH=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MCH}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle ACM rectangle en A, on sait que :
\begin{itemize}
\item AM=5,3 cm
\item \(\widehat{ACM}\) = 60\textdegree{}
\end{itemize}

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

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

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

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 Dans le triangle TNM rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TMN}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TMN}\)) = \(\dfrac{TN}{TM}\)
\par\addvspace{.5em}
d'où  tan(31\textdegree{}) =  \(\dfrac{3{,}4}{TM}\)
\par\addvspace{.5em}
On a donc TM = \(\dfrac{3{,}4}{\tan(31^{\circ})}\) \(\approx\) 5,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{LTF}\)) = \(\dfrac{LT}{FT}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{LTF}\)) =  \(\dfrac{5{,}7}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{LTF}\) = Arccos (\(\dfrac{5{,}7}{10}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{APG}\)) = \(\dfrac{AG}{GP}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{AG}{1{,}4}\)

On a donc AG = 1,4 \(\times\) sin(41\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MCH}\)) = \(\dfrac{MC}{CH}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MCH}\)) = \(\dfrac{2{,}5}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{MCH}\) =ArcCos(\(\dfrac{2{,}5}{7}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{ACM}\)) =  \(\dfrac{AM}{CM}\)
\par\addvspace{.5em}
 d'où sin(60\textdegree{}) = \(\dfrac{5{,}3}{CM}\)
\par\addvspace{.5em}
 On a donc CM = \(\dfrac{5{,}3}{\sin(60^{\circ})}\) \(\approx\) 6,1 cm\end{document}
