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\begin{document}

\sloppy

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

 Dans le triangle BCA rectangle en B, on sait que :
\begin{itemize}
\item BA=6,6 cm
\item CA=7,9 cm
\end{itemize}

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

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

 Dans le triangle PNH rectangle en P, on sait que :
\begin{itemize}
\item NH=0,4 cm
\item \(\widehat{NHP}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle SAD rectangle en S, on sait que :
\begin{itemize}
\item AD=4,9 cm
\item \(\widehat{SAD}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZAS rectangle en Z, on sait que :
\begin{itemize}
\item ZS=3,5 cm
\item \(\widehat{ASZ}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle HWV rectangle en H, on sait que :
\begin{itemize}
\item HW=2,8 cm
\item HV=5,8 cm
\end{itemize}

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

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

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

cos(\(\widehat{BAC}\)) = \(\dfrac{BA}{CA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BAC}\)) =  \(\dfrac{6{,}6}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{BAC}\) = Arccos (\(\dfrac{6{,}6}{7{,}9}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PHN}\)) = \(\dfrac{PH}{NH}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{PH}{0{,}4}\)

On a donc PH = 0,4 \(\times\) cos(26\textdegree{}) \(\approx\) 0{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{SAD}\)) = \(\dfrac{SA}{AD}\)
\par\addvspace{.5em}
d'où cos(75\textdegree{}) = \(\dfrac{SA}{4{,}9}\)

On a donc SA = 4,9 \(\times\) cos(75\textdegree{}) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ZSA}\)) = \(\dfrac{ZS}{AS}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{3{,}5}{AS}\)
\par\addvspace{.5em}
 On a donc AS =  \(\dfrac{3{,}5}{\cos(43^{\circ})}\) \(\approx\) 4,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.404,2.551) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HWV}\)) = \(\dfrac{HV}{HW}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HWV}\)) = \(\dfrac{5{,}8}{2{,}8}\)
\par\addvspace{.5em}
 On a donc HWV = ArcTan(\(\dfrac{5{,}8}{2{,}8}\)) \(\approx\) 64\textdegree{}\end{document}
