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\begin{document}

\sloppy

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

 Dans le triangle BPK rectangle en B, on sait que :
\begin{itemize}
\item BK=3,7 cm
\item PK=6,9 cm
\end{itemize}

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

 Dans le triangle SDR rectangle en S, on sait que :
\begin{itemize}
\item SR=6,3 cm
\item DR=7,7 cm
\end{itemize}

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

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

 Dans le triangle HPT rectangle en H, on sait que :
\begin{itemize}
\item HT=8,8 cm
\item \(\widehat{HPT}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAH rectangle en D, on sait que :
\begin{itemize}
\item DA=5,8 cm
\item \(\widehat{AHD}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle MCD rectangle en M, on sait que :
\begin{itemize}
\item CD=5,7 cm
\item \(\widehat{CDM}\) =  41\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{BPK}\)) =  \(\dfrac{BK}{PK}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BPK}\)) = \(\dfrac{3{,}7}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{BPK}\) =  ArcSin(\(\dfrac{3{,}7}{6{,}9}\)) \(\approx\) 32\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SRD}\)) = \(\dfrac{SR}{DR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SRD}\)) =  \(\dfrac{6{,}3}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SRD}\) = Arccos (\(\dfrac{6{,}3}{7{,}7}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HPT}\)) =  \(\dfrac{HT}{PT}\)
\par\addvspace{.5em}
 d'où sin(69\textdegree{}) = \(\dfrac{8{,}8}{PT}\)
\par\addvspace{.5em}
 On a donc PT = \(\dfrac{8{,}8}{\sin(69^{\circ})}\) \(\approx\) 9,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DHA}\)) = \(\dfrac{DA}{DH}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{5{,}8}{DH}\)
\par\addvspace{.5em}
On a donc DH = \(\dfrac{5{,}8}{\tan(40^{\circ})}\) \(\approx\) 6,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{MDC}\)) = \(\dfrac{MC}{CD}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{MC}{5{,}7}\)

On a donc MC = 5,7 \(\times\) sin(41\textdegree{}) \(\approx\) 3,7 cm\end{document}
