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\begin{document}

\sloppy

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

 Dans le triangle DRB rectangle en D, on sait que :
\begin{itemize}
\item DR=2,5 cm
\item RB=9,8 cm
\end{itemize}

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

 Dans le triangle KCT rectangle en K, on sait que :
\begin{itemize}
\item KC=0,6 cm
\item \(\widehat{KCT}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle RBC rectangle en R, on sait que :
\begin{itemize}
\item RB=2,1 cm
\item BC=7,2 cm
\end{itemize}

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

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

 Dans le triangle PWD rectangle en P, on sait que :
\begin{itemize}
\item PW=3,7 cm
\item \(\widehat{PWD}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWT rectangle en F, on sait que :
\begin{itemize}
\item WT=1,5 cm
\item \(\widehat{FWT}\) = 56\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [FW]. (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 (1.457,2.052) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DRB}\)) = \(\dfrac{DR}{RB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DRB}\)) = \(\dfrac{2{,}5}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{DRB}\) =ArcCos(\(\dfrac{2{,}5}{9{,}8}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KCT}\)) = \(\dfrac{KC}{CT}\)
\par\addvspace{.5em}
 d'où cos(48\textdegree{}) = \(\dfrac{0{,}6}{CT}\)
\par\addvspace{.5em}
On a donc CT = \(\dfrac{0{,}6}{\cos(48^{\circ})}\)\(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{RCB}\)) = \(\dfrac{RB}{BC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RCB}\)) =\(\dfrac{2{,}1}{7{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RCB}\) = ArcSin(\(\dfrac{2{,}1}{7{,}2}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{PWD}\)) = \(\dfrac{PD}{PW}\)
\par\addvspace{.5em}
d'où tan(76\textdegree{}) =\(\dfrac{PD}{3{,}7}\)
\par\addvspace{.5em}
On a donc PD = 3,7 \(\times\) tan(76\textdegree{}) \(\approx\) 14,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FWT}\)) = \(\dfrac{FW}{WT}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{FW}{1{,}5}\)

On a donc FW = 1,5 \(\times\) cos(56\textdegree{}) \(\approx\) 0,8 cm\end{document}
