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\begin{document}

\sloppy

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

 Dans le triangle PCN rectangle en P, on sait que :
\begin{itemize}
\item CN=1 cm
\item \(\widehat{CNP}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle AHS rectangle en A, on sait que :
\begin{itemize}
\item AH=4,7 cm
\item \(\widehat{HSA}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle SZA rectangle en S, on sait que :
\begin{itemize}
\item SA=1,6 cm
\item \(\widehat{SZA}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZPA rectangle en Z, on sait que :
\begin{itemize}
\item ZA=5,7 cm
\item PA=7,8 cm
\end{itemize}

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

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

 Dans le triangle MCT rectangle en M, on sait que :
\begin{itemize}
\item MT=6,2 cm
\item CT=8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MCT}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

cos(\(\widehat{PNC}\)) = \(\dfrac{PN}{CN}\)
\par\addvspace{.5em}
  d'où    cos(20\textdegree{}) =  \(\dfrac{PN}{1}\)

On a donc PN = 1 \(\times\) cos(20\textdegree{}) \(\approx\) 0{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ASH}\)) = \(\dfrac{AH}{HS}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{4{,}7}{HS}\)
\par\addvspace{.5em}
On a donc HS =\(\dfrac{4{,}7}{\sin(31^{\circ})}\) \(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{SZA}\)) = \(\dfrac{SA}{SZ}\)
\par\addvspace{.5em}
tan(76\textdegree{}) = \(\dfrac{1{,}6}{SZ}\)
\par\addvspace{.5em}
 On a donc SZ = \(\dfrac{1{,}6}{\tan(76^{\circ})}\) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZAP}\)) = \(\dfrac{ZA}{PA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZAP}\)) =  \(\dfrac{5{,}7}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZAP}\) = Arccos (\(\dfrac{5{,}7}{7{,}8}\)) \(\approx\) 43\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.069,2.731) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MCT}\)) =  \(\dfrac{MT}{CT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MCT}\)) = \(\dfrac{6{,}2}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{MCT}\) =  ArcSin(\(\dfrac{6{,}2}{8}\)) \(\approx\) 51\textdegree{}\end{document}
