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\begin{document}

\sloppy

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

 Dans le triangle DPG rectangle en D, on sait que :
\begin{itemize}
\item DP=1,8 cm
\item \(\widehat{PGD}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle KHR rectangle en K, on sait que :
\begin{itemize}
\item KR=6 cm
\item HR=9,1 cm
\end{itemize}

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

 Dans le triangle HGA rectangle en H, on sait que :
\begin{itemize}
\item HG=7 cm
\item \(\widehat{GAH}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle RLS rectangle en R, on sait que :
\begin{itemize}
\item RS=1,1 cm
\item \(\widehat{RLS}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle PWD rectangle en P, on sait que :
\begin{itemize}
\item PW=1,1 cm
\item PD=4,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DGP}\)) = \(\dfrac{DP}{DG}\)
\par\addvspace{.5em}
d'où  tan(30\textdegree{}) =  \(\dfrac{1{,}8}{DG}\)
\par\addvspace{.5em}
On a donc DG = \(\dfrac{1{,}8}{\tan(30^{\circ})}\) \(\approx\) 3,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KHR}\)) =  \(\dfrac{KR}{HR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KHR}\)) = \(\dfrac{6}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{KHR}\) =  ArcSin(\(\dfrac{6}{9{,}1}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HAG}\)) = \(\dfrac{HG}{GA}\)
\par\addvspace{.5em}
 D'où sin(16\textdegree{}) = \(\dfrac{7}{GA}\)
\par\addvspace{.5em}
On a donc GA =\(\dfrac{7}{\sin(16^{\circ})}\) \(\approx\) 25,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RLS}\)) = \(\dfrac{RS}{RL}\)
\par\addvspace{.5em}
tan(66\textdegree{}) = \(\dfrac{1{,}1}{RL}\)
\par\addvspace{.5em}
 On a donc RL = \(\dfrac{1{,}1}{\tan(66^{\circ})}\) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.068) {4,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.414) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PDW}\)) = \(\dfrac{PW}{PD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{PDW}\)) =  \(\dfrac{1{,}1}{4{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PDW}\) =  ArcTan(\(\dfrac{1{,}1}{4{,}5}\)) \(\approx\) 14\textdegree{}\end{document}
