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\begin{document}

\sloppy

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

 Dans le triangle SFP rectangle en S, on sait que :
\begin{itemize}
\item SP=1,1 cm
\item \(\widehat{FPS}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJN rectangle en M, on sait que :
\begin{itemize}
\item MN=0,8 cm
\item \(\widehat{MJN}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle MAS rectangle en M, on sait que :
\begin{itemize}
\item MS=5,1 cm
\item AS=8,2 cm
\end{itemize}

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

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

 Dans le triangle NHP rectangle en N, on sait que :
\begin{itemize}
\item NH=7,2 cm
\item \(\widehat{HPN}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle PAB rectangle en P, on sait que :
\begin{itemize}
\item PA=2,2 cm
\item AB=9,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PAB}\). \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 (4.284,0.788) {45°};
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 Dans le triangle SFP rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SPF}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{SPF}\)) = \(\dfrac{SP}{FP}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{1{,}1}{FP}\)
\par\addvspace{.5em}
 On a donc FP =  \(\dfrac{1{,}1}{\cos(45^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MJN}\)) = \(\dfrac{MN}{MJ}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{0{,}8}{MJ}\)
\par\addvspace{.5em}
 On a donc MJ = \(\dfrac{0{,}8}{\tan(77^{\circ})}\) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{MSA}\)) = \(\dfrac{MS}{AS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MSA}\)) =  \(\dfrac{5{,}1}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{MSA}\) = Arccos (\(\dfrac{5{,}1}{8{,}2}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NPH}\)) = \(\dfrac{NH}{NP}\)
\par\addvspace{.5em}
d'où  tan(43\textdegree{}) =  \(\dfrac{7{,}2}{NP}\)
\par\addvspace{.5em}
On a donc NP = \(\dfrac{7{,}2}{\tan(43^{\circ})}\) \(\approx\) 7,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.466,1.971) {?};
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 Dans le triangle PAB rectangle en P, on cherche une relation entre l'angle aigu   \(\widehat{PAB}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{PAB}\)) = \(\dfrac{PA}{AB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PAB}\)) = \(\dfrac{2{,}2}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{PAB}\) =ArcCos(\(\dfrac{2{,}2}{9{,}9}\)) \(\approx\)77\textdegree{}\end{document}
