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\begin{document}

\sloppy

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

 Dans le triangle DCS rectangle en D, on sait que :
\begin{itemize}
\item DC=2,6 cm
\item DS=4,3 cm
\end{itemize}

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

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

 Dans le triangle BGR rectangle en B, on sait que :
\begin{itemize}
\item BG=6,3 cm
\item \(\widehat{BGR}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle SAP rectangle en S, on sait que :
\begin{itemize}
\item SP=0,6 cm
\item \(\widehat{SAP}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle BTP rectangle en B, on sait que :
\begin{itemize}
\item BP=6,2 cm
\item TP=10 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BTP}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle MTV rectangle en M, on sait que :
\begin{itemize}
\item MT=9,9 cm
\item \(\widehat{MTV}\) = 53\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DSC}\)) = \(\dfrac{DC}{DS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DSC}\)) =  \(\dfrac{2{,}6}{4{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DSC}\) =  ArcTan(\(\dfrac{2{,}6}{4{,}3}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{BGR}\)) = \(\dfrac{BR}{BG}\)
\par\addvspace{.5em}
d'où tan(47\textdegree{}) =\(\dfrac{BR}{6{,}3}\)
\par\addvspace{.5em}
On a donc BR = 6,3 \(\times\) tan(47\textdegree{}) \(\approx\) 6,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{SAP}\)) = \(\dfrac{SP}{SA}\)
\par\addvspace{.5em}
tan(75\textdegree{}) = \(\dfrac{0{,}6}{SA}\)
\par\addvspace{.5em}
 On a donc SA = \(\dfrac{0{,}6}{\tan(75^{\circ})}\) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BTP}\)) =  \(\dfrac{BP}{TP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BTP}\)) = \(\dfrac{6{,}2}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{BTP}\) =  ArcSin(\(\dfrac{6{,}2}{10}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.940,2.737) {53°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MTV}\)) = \(\dfrac{MT}{TV}\)
\par\addvspace{.5em}
 d'où cos(53\textdegree{}) = \(\dfrac{9{,}9}{TV}\)
\par\addvspace{.5em}
On a donc TV = \(\dfrac{9{,}9}{\cos(53^{\circ})}\)\(\approx\) 16,5 cm\end{document}
