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\begin{document}

\sloppy

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

 Dans le triangle WNP rectangle en W, on sait que :
\begin{itemize}
\item WN=7,7 cm
\item \(\widehat{NPW}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle AVM rectangle en A, on sait que :
\begin{itemize}
\item AV=2,4 cm
\item AM=5,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AVM}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MJZ rectangle en M, on sait que :
\begin{itemize}
\item JZ=4 cm
\item \(\widehat{MJZ}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZHP rectangle en Z, on sait que :
\begin{itemize}
\item ZH=9,3 cm
\item \(\widehat{ZHP}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle PWS rectangle en P, on sait que :
\begin{itemize}
\item PW=2,3 cm
\item WS=9,4 cm
\end{itemize}

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

\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 (4.864,0.734) {35°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WPN}\)) = \(\dfrac{WN}{WP}\)
\par\addvspace{.5em}
d'où  tan(35\textdegree{}) =  \(\dfrac{7{,}7}{WP}\)
\par\addvspace{.5em}
On a donc WP = \(\dfrac{7{,}7}{\tan(35^{\circ})}\) \(\approx\) 11 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{AVM}\)) = \(\dfrac{AM}{AV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{AVM}\)) = \(\dfrac{5{,}9}{2{,}4}\)
\par\addvspace{.5em}
 On a donc AVM = ArcTan(\(\dfrac{5{,}9}{2{,}4}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MJZ}\)) = \(\dfrac{MJ}{JZ}\)
\par\addvspace{.5em}
d'où cos(74\textdegree{}) = \(\dfrac{MJ}{4}\)

On a donc MJ = 4 \(\times\) cos(74\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZHP}\)) = \(\dfrac{ZH}{HP}\)
\par\addvspace{.5em}
 d'où cos(61\textdegree{}) = \(\dfrac{9{,}3}{HP}\)
\par\addvspace{.5em}
On a donc HP = \(\dfrac{9{,}3}{\cos(61^{\circ})}\)\(\approx\) 19,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \draw[black] (1.050,1.505) -- (1.092,1.506) -- (1.134,1.510) -- (1.174,1.505);
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.248) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.397) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PSW}\)) = \(\dfrac{PW}{WS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PSW}\)) =\(\dfrac{2{,}3}{9{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PSW}\) = ArcSin(\(\dfrac{2{,}3}{9{,}4}\)) \(\approx\) 14\textdegree{}\end{document}
