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\begin{document}

\sloppy

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

 Dans le triangle DBC rectangle en D, on sait que :
\begin{itemize}
\item DB=2,2 cm
\item DC=3,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DBC}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle WBZ rectangle en W, on sait que :
\begin{itemize}
\item WB=3,5 cm
\item \(\widehat{BZW}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle CGK rectangle en C, on sait que :
\begin{itemize}
\item CG=1,5 cm
\item GK=10 cm
\end{itemize}

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

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

 Dans le triangle APR rectangle en A, on sait que :
\begin{itemize}
\item AR=2,6 cm
\item \(\widehat{APR}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle CJG rectangle en C, on sait que :
\begin{itemize}
\item JG=7 cm
\item \(\widehat{CJG}\) = 76\textdegree{}
\end{itemize}

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

\newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{DBC}\)) = \(\dfrac{DC}{DB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DBC}\)) = \(\dfrac{3{,}9}{2{,}2}\)
\par\addvspace{.5em}
 On a donc DBC = ArcTan(\(\dfrac{3{,}9}{2{,}2}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WZB}\)) = \(\dfrac{WB}{WZ}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{3{,}5}{WZ}\)
\par\addvspace{.5em}
On a donc WZ = \(\dfrac{3{,}5}{\tan(19^{\circ})}\) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.285,1.612) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CKG}\)) = \(\dfrac{CG}{GK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CKG}\)) =\(\dfrac{1{,}5}{10}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CKG}\) = ArcSin(\(\dfrac{1{,}5}{10}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{APR}\)) =  \(\dfrac{AR}{PR}\)
\par\addvspace{.5em}
 d'où sin(63\textdegree{}) = \(\dfrac{2{,}6}{PR}\)
\par\addvspace{.5em}
 On a donc PR = \(\dfrac{2{,}6}{\sin(63^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.256) {G};
  \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) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.460,2.019) {76°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{CJG}\)) = \(\dfrac{CJ}{JG}\)
\par\addvspace{.5em}
d'où cos(76\textdegree{}) = \(\dfrac{CJ}{7}\)

On a donc CJ = 7 \(\times\) cos(76\textdegree{}) \(\approx\) 1,7 cm\end{document}
