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\begin{document}

\sloppy

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

 Dans le triangle WGA rectangle en W, on sait que :
\begin{itemize}
\item WA=8,3 cm
\item \(\widehat{GAW}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle THV rectangle en T, on sait que :
\begin{itemize}
\item TH=1,9 cm
\item TV=4,9 cm
\end{itemize}

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

 Dans le triangle CPN rectangle en C, on sait que :
\begin{itemize}
\item PN=5,4 cm
\item \(\widehat{PNC}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle PJM rectangle en P, on sait que :
\begin{itemize}
\item PJ=2,4 cm
\item JM=8 cm
\end{itemize}

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

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

 Dans le triangle ZBL rectangle en Z, on sait que :
\begin{itemize}
\item BL=3,6 cm
\item \(\widehat{BLZ}\) =  21\textdegree{}
\end{itemize}

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

\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 (3.500,1.146) {8,3 cm};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.482) {12°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WAG}\)) = \(\dfrac{WA}{GA}\)
\par\addvspace{.5em}
  d'où    cos(12\textdegree{}) =  \(\dfrac{8{,}3}{GA}\)
\par\addvspace{.5em}
 On a donc GA =  \(\dfrac{8{,}3}{\cos(12^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle THV rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{THV}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{THV}\)) = \(\dfrac{TV}{TH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{THV}\)) = \(\dfrac{4{,}9}{1{,}9}\)
\par\addvspace{.5em}
 On a donc THV = ArcTan(\(\dfrac{4{,}9}{1{,}9}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CNP}\)) = \(\dfrac{CN}{PN}\)
\par\addvspace{.5em}
  d'où    cos(30\textdegree{}) =  \(\dfrac{CN}{5{,}4}\)

On a donc CN = 5,4 \(\times\) cos(30\textdegree{}) \(\approx\) 4{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PMJ}\)) = \(\dfrac{PJ}{JM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{PMJ}\)) =\(\dfrac{2{,}4}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PMJ}\) = ArcSin(\(\dfrac{2{,}4}{8}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZLB}\)) = \(\dfrac{ZB}{BL}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{ZB}{3{,}6}\)

On a donc ZB = 3,6 \(\times\) sin(21\textdegree{}) \(\approx\) 1,3 cm\end{document}
