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\begin{document}

\sloppy

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

 Dans le triangle GFM rectangle en G, on sait que :
\begin{itemize}
\item GM=9,8 cm
\item \(\widehat{FMG}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle LPA rectangle en L, on sait que :
\begin{itemize}
\item LP=1,8 cm
\item \(\widehat{PAL}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle WBK rectangle en W, on sait que :
\begin{itemize}
\item WK=7,1 cm
\item \(\widehat{WBK}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle NMF rectangle en N, on sait que :
\begin{itemize}
\item NM=1 cm
\item MF=7 cm
\end{itemize}

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

 Dans le triangle HDN rectangle en H, on sait que :
\begin{itemize}
\item HD=1 cm
\item DN=10 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{GMF}\)) = \(\dfrac{GM}{FM}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{9{,}8}{FM}\)
\par\addvspace{.5em}
 On a donc FM =  \(\dfrac{9{,}8}{\cos(29^{\circ})}\) \(\approx\) 11,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LAP}\)) = \(\dfrac{LP}{LA}\)
\par\addvspace{.5em}
d'où  tan(41\textdegree{}) =  \(\dfrac{1{,}8}{LA}\)
\par\addvspace{.5em}
On a donc LA = \(\dfrac{1{,}8}{\tan(41^{\circ})}\) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{WBK}\)) = \(\dfrac{WK}{WB}\)
\par\addvspace{.5em}
tan(47\textdegree{}) = \(\dfrac{7{,}1}{WB}\)
\par\addvspace{.5em}
 On a donc WB = \(\dfrac{7{,}1}{\tan(47^{\circ})}\) \(\approx\) 6,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NMF}\)) = \(\dfrac{NM}{MF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NMF}\)) = \(\dfrac{1}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{NMF}\) =ArcCos(\(\dfrac{1}{7}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{HND}\)) = \(\dfrac{HD}{DN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HND}\)) =\(\dfrac{1}{10}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HND}\) = ArcSin(\(\dfrac{1}{10}\)) \(\approx\) 6\textdegree{}\end{document}
